Further complex numbers

Further complex numbers

Introduction This chapter extends on what you have learnt in FP1
You will learn how to find the complex roots of numbers You will learn how to use De Moivre’s theorem in solving equations You will see how to plot the loci of points following a rule on an Argand diagram You will see how to solve problems involving transforming a set of values in one plane into another plane

Teachings for exercise 3A

𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) Further complex numbers y You can express a complex number in the form z = r(cosθ + isinθ) You should hopefully remember the modulus-argument form of a complex number z = x + iy from FP1 The value r is the modulus of the complex number, its distance from the origin (0,0) The argument is the angle the complex number makes with the positive x-axis, where: -π < θ ≤ π To show this visually… z (x,y) r y θ x x r is the modulus of z, its absolute value  This can be calculated using Pythagoras’ Theorem By GCSE trigonometry, length x = rcosθ and length y = rsinθ 𝑥=𝑟𝑐𝑜𝑠𝜃 𝑦=𝑟𝑠𝑖𝑛𝜃 𝑟= 𝑧 = 𝑥 2 + 𝑦 2 𝑧=𝑥+𝑖𝑦 Replace x and y using the values above 𝑧=𝑟𝑐𝑜𝑠𝜃 + 𝑖𝑟𝑠𝑖𝑛𝜃 Factorise by taking out r 𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 3A

𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) Further complex numbers y You can express a complex number in the form z = r(cosθ + isinθ) Express the following complex number in the modulus-argument form: z = -√3 + i To do this you need to find both the argument and the modulus of the complex number  Start by sketching it on an Argand diagram This is the argument Pay attention to the directions The ‘x’ part is negative so will go in the negative direction horizontally The ‘y’ part is positive so will go upwards r 1 θ x √3  Once sketched you can then find the modulus and argument using GCSE Pythagoras and Trigonometry 𝑟= (1) 2 𝑇𝑎𝑛𝜃= 1 3 Calculate Inverse Tan 𝑟=2 𝜃= 𝜋 6 𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) Remember that the argument is measured from the positive x-axis! Replace r and θ Subtract from π 𝑧=2 𝑐𝑜𝑠 5𝜋 6 +𝑖𝑠𝑖𝑛 5𝜋 6 arg 𝑧= 5𝜋 6 3A

𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) Further complex numbers You can express a complex number in the form z = r(cosθ + isinθ) Express the following complex number in the modulus-argument form: z = -√3 + i To do this you need to find both the argument and the modulus of the complex number  Start by sketching it on an Argand diagram 𝑧=2 𝑐𝑜𝑠 5𝜋 6 +𝑖𝑠𝑖𝑛 5𝜋 6 Remember that the argument is not unique We could add 2π to them and the result would be the same, because 2π radians is a complete turn 3A

𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) Further complex numbers y You can express a complex number in the form z = r(cosθ + isinθ) Express the following complex number in the modulus-argument form: z = 1 - i To do this you need to find both the argument and the modulus of the complex number  Start by sketching it on an Argand diagram Pay attention to the directions The ‘x’ part is positive so will go in the positive direction horizontally The ‘y’ part is negative so will go downwards 1 x θ 1 r  Once sketched you can then find the modulus and argument using GCSE Pythagoras and Trigonometry 𝑇𝑎𝑛𝜃= 1 1 𝑟= (1) 2 Calculate Inverse Tan 𝑟= 2 𝜃= 𝜋 4 𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) Negative as below the x-axis Replace r and θ 𝑧= 2 𝑐𝑜𝑠 − 𝜋 4 +𝑖𝑠𝑖𝑛 − 𝜋 4 arg 𝑧=− 𝜋 4 3A

𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) Further complex numbers 𝑧= 𝑟𝑒 𝑖𝜃 You can express a complex number in the form z = reiθ In chapter 6 you will meet series expansions of cosθ and sinθ This can be used to prove the following result (which we will do when we come to chapter 6) If z = x + iy then the complex number can also be written in this way: z = reiθ As before, r is the modulus of the complex number and θ is the argument  This form is known as the ‘exponential form’ 3A

𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) Further complex numbers 𝑧= 𝑟𝑒 𝑖𝜃 y You can express a complex number in the form z = reiθ Express the following complex number in the form reiθ, where -π < θ ≤ π z = 2 – 3i As with the modulus-argument form, you should start by sketching an Argand diagram and use it to find r and θ Pay attention to the directions The ‘x’ part is positive so will go in the positive direction horizontally The ‘y’ part is negative so will go downwards 2 x θ 3 r  Once sketched you can then find the modulus and argument using GCSE Pythagoras and Trigonometry 𝑟= (3) 2 𝑇𝑎𝑛𝜃= 3 2 Calculate Inverse Tan 𝑟= 13 𝑧= 𝑟𝑒 𝑖𝜃 𝜃=0.98 Replace r and θ Negative as below the x-axis 𝑧= 13 𝑒 −0.98𝑖 arg 𝑧=−0.98 3A

𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑐𝑜𝑠 −𝜃 =𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 −𝜃 =−𝑠𝑖𝑛𝜃 Further complex numbers 𝑧= 𝑟𝑒 𝑖𝜃 y y = cosθ You can express a complex number in the form z = reiθ In Core 2, you will have seen the following: cos(-θ) = cosθ sin(-θ) = -sinθ 1 -θ θ θ -360º -270º -180º -90º -θ θ 90º 180º 270º -1 You can see that cos(-θ) = cosθ anywhere on the graph y y = sinθ -90º -270º -360º -180º 1 -θ θ θ 90º 180º 270º -1 You can see that sin(-θ) = -sinθ anywhere on the graph 3A

𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑐𝑜𝑠 −𝜃 =𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 −𝜃 =−𝑠𝑖𝑛𝜃 Further complex numbers 𝑧= 𝑟𝑒 𝑖𝜃 𝑧= 2 𝑐𝑜𝑠 𝜋 10 +𝑖𝑠𝑖𝑛 𝜋 10 You can express a complex number in the form z = reiθ Express the following in the form z = reiθ where –π < θ ≤ π You can see from the form that r = √2 You can see from the form that θ = π/10 𝑧= 2 𝑐𝑜𝑠 𝜋 10 +𝑖𝑠𝑖𝑛 𝜋 10 𝜃= 𝜋 10 𝑟= 2 𝑧= 𝑟𝑒 𝑖𝜃 Replace r and θ 𝑧= 2 𝑒 𝜋 10 𝑖 3A

𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑐𝑜𝑠 −𝜃 =𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 −𝜃 =−𝑠𝑖𝑛𝜃 Further complex numbers 𝑧= 𝑟𝑒 𝑖𝜃 You can express a complex number in the form z = reiθ Express the following in the form z = reiθ where –π < θ ≤ π 𝑧=5 𝑐𝑜𝑠 𝜋 8 −𝑖𝑠𝑖𝑛 𝜋 8 𝑧=5 𝑐𝑜𝑠 𝜋 8 −𝑖𝑠𝑖𝑛 𝜋 8 We need to adjust this first The sign in the centre is negative, we need it to be positive for the ‘rules’ to work We also need both angles to be identical. In this case we can apply the rules we saw a moment ago… 3A

𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑐𝑜𝑠 −𝜃 =𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 −𝜃 =−𝑠𝑖𝑛𝜃 Further complex numbers 𝑧= 𝑟𝑒 𝑖𝜃 You can express a complex number in the form z = reiθ Express the following in the form z = reiθ where –π < θ ≤ π 𝑧=5 𝑐𝑜𝑠 𝜋 8 −𝑖𝑠𝑖𝑛 𝜋 8 Apply cosθ = cos(-θ) Apply sin(-θ) = -sin(θ) 𝑧=5 𝑐𝑜𝑠 − 𝜋 8 + 𝑖𝑠𝑖𝑛 − 𝜋 8 𝑧=5 𝑐𝑜𝑠 𝜋 8 −𝑖𝑠𝑖𝑛 𝜋 8 You can see from the form that r = 5 You can see from the form that θ = -π/8 𝑟=5 𝜃=− 𝜋 8 𝑧= 𝑟𝑒 𝑖𝜃 Replace r and θ 𝑧= 5𝑒 − 𝜋 8 𝑖 3A

𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑐𝑜𝑠 −𝜃 =𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 −𝜃 =−𝑠𝑖𝑛𝜃 Further complex numbers 𝑧= 𝑟𝑒 𝑖𝜃 𝑧= 2 𝑒 3𝜋 4 𝑖 You can express a complex number in the form z = reiθ Express the following in the form z = x + iy where 𝑥∈ℝ and 𝑦∈ℝ You can see from the form that r = √2 You can see from the form that θ = 3π/4 𝑧= 2 𝑒 3𝜋 4 𝑖 𝑟= 2 𝜃= 3𝜋 4 𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) Replace r and θ This means that x and y have to be real numbers (ie not complex) 𝑧= 2 𝑐𝑜𝑠 3𝜋 4 +𝑖𝑠𝑖𝑛 3𝜋 4 You can calculate all of this! Leave the second part in terms of i 𝑧=−1+𝑖 3A

𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑐𝑜𝑠 −𝜃 =𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 −𝜃 =−𝑠𝑖𝑛𝜃 Further complex numbers 𝑧= 𝑟𝑒 𝑖𝜃 𝑧=2 𝑒 23𝜋 5 𝑖 You can express a complex number in the form z = reiθ Express the following in the form r(cosθ + isinθ), where –π < θ ≤ π The value of θ is not in the range we want. We can keep subtracting 2π until it is! You can see from the form that r = 2 You can see from the form that θ = 23π/5 𝑧=2 𝑒 23𝜋 5 𝑖 𝑟=2 𝜃= 23𝜋 5 Subtract 2π 𝜃= 13𝜋 5 Subtract 2π 𝜃= 3𝜋 5 𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) Replace r and θ 𝑧=2 𝑐𝑜𝑠 3𝜋 5 +𝑖𝑠𝑖𝑛 3𝜋 5 3A

𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑐𝑜𝑠 −𝜃 =𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 −𝜃 =−𝑠𝑖𝑛𝜃 Further complex numbers 𝑧= 𝑟𝑒 𝑖𝜃 You can express a complex number in the form z = reiθ Use: To show that: 𝑒 𝑖𝜃 =𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 Let θ = -θ 𝑒 𝑖(−𝜃) =𝑐𝑜𝑠 −𝜃 +𝑖𝑠𝑖𝑛 −𝜃 Use the relationships above to rewrite 𝑒 −𝑖𝜃 =𝑐𝑜𝑠𝜃−𝑖𝑠𝑖𝑛𝜃 𝑒 𝑖𝜃 =𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 1) 𝑒 𝑖𝜃 =𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃= 𝑒 𝑖𝜃 + 𝑒 −𝑖𝜃 2) 𝑒 −𝑖𝜃 =𝑐𝑜𝑠𝜃−𝑖𝑠𝑖𝑛𝜃 Add 1 and 2 𝑒 −𝑖𝜃 + 𝑒 𝑖𝜃 =2𝑐𝑜𝑠𝜃 Divide by 2 1 2 𝑒 −𝑖𝜃 + 𝑒 𝑖𝜃 =𝑐𝑜𝑠𝜃 3A

Teachings for exercise 3B

𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2 Further complex numbers Multiplying a complex number z1 by another complex number z2, both in the modulus-argument form You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number To be able to do this you need to be able to use the identities for sine and cosine of two angles added or subtracted – you will have seen these in Core 3 𝑧 1 = 𝑟 1 𝑐𝑜𝑠 𝜃 1 +𝑖𝑠𝑖𝑛 𝜃 1 𝑧 2 = 𝑟 2 𝑐𝑜𝑠 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑐𝑜𝑠 𝜃 1 +𝑖𝑠𝑖𝑛 𝜃 1 × 𝑟 2 𝑐𝑜𝑠 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2 Rewrite 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 +𝑖𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2 Now you can expand the double bracket as you would with a quadratic 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 +𝑖𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 + 𝑖 2 𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2 𝑠𝑖𝑛 𝜃 1 ± 𝜃 2 =𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 ±𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 +𝑖𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 −𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2 Group terms using the identities to the left  You can also factorise the ‘i’ out 𝑐𝑜𝑠 𝜃 1 ± 𝜃 2 =𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 ∓𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 2 𝜃+ 𝑠𝑖𝑛 2 𝜃=1 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 + 𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2 So when multiplying two complex numbers in the modulus-argument form: Multiply the moduli Add the arguments together The form of the answer is the same 3B

𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 ) Further complex numbers Multiplying a complex number z1 by another complex number z2, both in the exponential form You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number To be able to do this you need to be able to use the identities for sine and cosine of two angles added or subtracted – you will have seen these in Core 3 𝑧 1 = 𝑟 1 𝑒 𝑖 𝜃 1 𝑧 2 = 𝑟 2 𝑒 𝑖 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑒 𝑖 𝜃 1 𝑟 2 𝑒 𝑖 𝜃 2 Rewrite  Remember you add the powers in this situation 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 𝜃 1 + 𝑖𝜃 2 You can factorise the power 𝑠𝑖𝑛 𝜃 1 ± 𝜃 2 =𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 ±𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 ) 𝑐𝑜𝑠 𝜃 1 ± 𝜃 2 =𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 ∓𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2 You can see that in this form the process is essentially the same as for the modulus-argument form: Multiply the moduli together Add the arguments together The answer is in the same form 𝑐𝑜𝑠 2 𝜃+ 𝑠𝑖𝑛 2 𝜃=1 3B

𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 ) 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2 Further complex numbers Dividing a complex number z1 by another complex number z2, both in the modulus-argument form You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number To be able to do this you need to be able to use the identities for sine and cosine of two angles added or subtracted – you will have seen these in Core 3 𝑧 1 = 𝑟 1 𝑐𝑜𝑠 𝜃 1 +𝑖𝑠𝑖𝑛 𝜃 1 𝑧 2 = 𝑟 2 𝑐𝑜𝑠 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑐𝑜𝑠 𝜃 1 +𝑖𝑠𝑖𝑛 𝜃 1 𝑟 2 𝑐𝑜𝑠 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2 Multiply to cancel terms on the denominator 𝑧 1 𝑧 2 = 𝑟 1 𝑐𝑜𝑠 𝜃 1 +𝑖𝑠𝑖𝑛 𝜃 1 𝑟 2 𝑐𝑜𝑠 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2 × 𝑐𝑜𝑠 𝜃 2 −𝑖𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 2 −𝑖𝑠𝑖𝑛 𝜃 2 Multiply out 𝑧 1 𝑧 2 = 𝑟 1 𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 −𝑖𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 − 𝑖 2 𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2 𝑟 2 𝑐𝑜𝑠 𝜃 2 𝑐𝑜𝑠 𝜃 2 −𝑖𝑐𝑜𝑠 𝜃 2 𝑠𝑖𝑛 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 2 − 𝑖 2 𝑠𝑖𝑛 𝜃 2 𝑠𝑖𝑛 𝜃 2 Remove i2 𝑠𝑖𝑛 𝜃 1 ± 𝜃 2 =𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 ±𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 −𝑖𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 +𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2 𝑟 2 𝑐𝑜𝑠 𝜃 2 𝑐𝑜𝑠 𝜃 2 −𝑖𝑐𝑜𝑠 𝜃 2 𝑠𝑖𝑛 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 2 +𝑠𝑖𝑛 𝜃 2 𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 1 ± 𝜃 2 =𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 ∓𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2 Group real and complex 𝑧 1 𝑧 2 = 𝑟 𝑟 2 𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 +𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2 + 𝑖 𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 −𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 2 𝜃+ 𝑠𝑖𝑛 2 𝜃=1 𝑐𝑜𝑠 2 𝜃 2 + 𝑠𝑖𝑛 2 𝜃 2 Rewrite terms So when dividing two complex numbers in the modulus-argument form: Divide the moduli Subtract the arguments The form of the answer is the same 𝑧 1 𝑧 2 = 𝑟 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 + 𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2 Rewrite (again!) 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2 3B

𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 ) 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 (𝜃 1 − 𝜃 2 ) Further complex numbers Dividing a complex number z1 by another complex number z2, both in the exponential form You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number To be able to do this you need to be able to use the identities for sine and cosine of two angles added or subtracted – you will have seen these in Core 3 𝑧 1 = 𝑟 1 𝑒 𝑖 𝜃 1 𝑧 2 = 𝑟 2 𝑒 𝑖 𝜃 2 𝑟 1 𝑒 𝑖 𝜃 1 𝑟 2 𝑒 𝑖 𝜃 2 𝑧 1 𝑧 2 = Rewrite terms  The denominator can be written with a negative power 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 𝜃 1 𝑒 −𝑖 𝜃 2 Multiplying so add the powers 𝑠𝑖𝑛 𝜃 1 ± 𝜃 2 =𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 ±𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 𝜃 1 −𝑖 𝜃 2 𝑐𝑜𝑠 𝜃 1 ± 𝜃 2 =𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 ∓𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2 Factorise the power 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 (𝜃 1 − 𝜃 2 ) 𝑐𝑜𝑠 2 𝜃+ 𝑠𝑖𝑛 2 𝜃=1 You can see that in this form the process is essentially the same as for the modulus-argument form: Divide the moduli Subtract the arguments The answer is in the same form 3B

𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 ) 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 (𝜃 1 − 𝜃 2 ) Further complex numbers 3 𝑐𝑜𝑠 5𝜋 12 +𝑖𝑠𝑖𝑛 5𝜋 12 ×4 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12 You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number Express the following calculation in the form x + iy: Combine using one of the rules above Multiply the moduli Add the arguments 3(4) 𝑐𝑜𝑠 5𝜋 12 + 𝜋 12 +𝑖𝑠𝑖𝑛 5𝜋 12 + 𝜋 12 Simplify terms 12 𝑐𝑜𝑠 𝜋 2 +𝑖𝑠𝑖𝑛 𝜋 2 3 𝑐𝑜𝑠 5𝜋 12 +𝑖𝑠𝑖𝑛 5𝜋 12 ×4 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12 Calculate the cos and sin parts (in terms of i where needed) 12 0+𝑖(1) Multiply out =12𝑖 3B

𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 ) 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 (𝜃 1 − 𝜃 2 ) Further complex numbers 2 𝑐𝑜𝑠 𝜋 15 +𝑖𝑠𝑖𝑛 𝜋 15 ×3 𝑐𝑜𝑠 2𝜋 5 −𝑖𝑠𝑖𝑛 2𝜋 5 You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number Express the following calculation in the form x + iy: cos(-θ) = cosθ sin(-θ) = -sinθ The cos and sin terms must be added for this to work!  Rewrite using the rules you saw in 3A 2 𝑐𝑜𝑠 𝜋 15 +𝑖𝑠𝑖𝑛 𝜋 15 ×3 𝑐𝑜𝑠 − 2𝜋 5 +𝑖𝑠𝑖𝑛 − 2𝜋 5 Combine using a rule from above 2(3) 𝑐𝑜𝑠 𝜋 15 − 2𝜋 5 +𝑖𝑠𝑖𝑛 𝜋 15 − 2𝜋 5 2 𝑐𝑜𝑠 𝜋 15 +𝑖𝑠𝑖𝑛 𝜋 15 ×3 𝑐𝑜𝑠 2𝜋 5 −𝑖𝑠𝑖𝑛 2𝜋 5 Simplify 6 𝑐𝑜𝑠 − 𝜋 3 +𝑖𝑠𝑖𝑛 − 𝜋 3 Calculate the cos and sin parts 𝑖 − Multiply out =3−3 3 𝑖 3B

𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 + 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 + 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 𝜃 2 ) 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 (𝜃 1 − 𝜃 2 ) Further complex numbers 2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 𝑐𝑜𝑠 5𝜋 6 +𝑖𝑠𝑖𝑛 5𝜋 6 You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number Express the following calculation in the form x + iy: Combine using one of the rules above Divide the moduli Subtract the arguments 2 2 𝑐𝑜𝑠 𝜋 12 − 5𝜋 6 +𝑖𝑠𝑖𝑛 𝜋 12 − 5𝜋 6 Simplify 2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 𝑐𝑜𝑠 5𝜋 6 +𝑖𝑠𝑖𝑛 5𝜋 6 2 2 𝑐𝑜𝑠 − 3𝜋 4 +𝑖𝑠𝑖𝑛 − 3𝜋 4 You can work out the sin and cos parts 2 2 − 𝑖 − 1 2 Multiply out =− 1 2 − 1 2 𝑖 3B

Teachings for exercise 3C

You need to be able to prove that [r(cosθ + isinθ)]n = rn(cos(nθ + isinnθ) for any integer n Let: z = r(cosθ + isinθ) 𝑧=𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑧 2 =𝑧×𝑧 Use the modulus-argument form 𝑧 2 =𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃)×𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) Multiply the moduli, add the arguments 𝑧 2 = 𝑟 2 (𝑐𝑜𝑠2𝜃+𝑖𝑠𝑖𝑛2𝜃) 𝑧= 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 1 =𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑧 3 = 𝑧 2 ×𝑧 Use the modulus-argument form 𝑧 2 = 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 2 = 𝑟 2 (𝑐𝑜𝑠2𝜃+𝑖𝑠𝑖𝑛2𝜃) 𝑧 3 = 𝑟 2 (𝑐𝑜𝑠2𝜃+𝑖𝑠𝑖𝑛2𝜃)×𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) Multiply the moduli, add the arguments 𝑧 3 = 𝑟 3 (𝑐𝑜𝑠3𝜃+𝑖𝑠𝑖𝑛3𝜃) 𝑧 3 = 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 3 = 𝑟 3 (𝑐𝑜𝑠3𝜃+𝑖𝑠𝑖𝑛3𝜃) 𝑧 4 = 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 4 = 𝑟 4 (𝑐𝑜𝑠4𝜃+𝑖𝑠𝑖𝑛4𝜃) 𝑧 4 = 𝑧 3 ×𝑧 Use the modulus-argument form 𝑧 4 = 𝑟 3 (𝑐𝑜𝑠3𝜃+𝑖𝑠𝑖𝑛3𝜃)×𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃) Multiply the moduli, add the arguments 𝑧 4 = 𝑟 4 (𝑐𝑜𝑠4𝜃+𝑖𝑠𝑖𝑛4𝜃) This is De Moivre’s Theorem  You need to be able to prove this De Moivre = ‘De Mwavre’ (pronunciation) 3C

𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃) Further complex numbers Proving that: 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 You need to be able to prove that [r(cosθ + isinθ)]n = rn(cos(nθ + isinnθ) for any integer n De Moivre’s theorem can be proved using the method of proof by induction from FP1 Basis – show the statement is true for n = 1 Assumption – assume the statement is true for n = k Inductive – show that if true for n = k, then the statement is also true for n = k + 1 Conclusion – because the statement is true for n = 1 and also true if any value is, then the statement is true for all values of n is true for all positive integers BASIS  Show that the statement is true for n = 1 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 Sub in n = 1 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 1 = 𝑟 1 𝑐𝑜𝑠1𝜃+𝑖𝑠𝑖𝑛1𝜃 Simplify each side 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 =𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 ASSUMPTION  Assume that the statement is true for n = k 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 Replace n with k 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑘 = 𝑟 𝑘 𝑐𝑜𝑠𝑘𝜃+𝑖𝑠𝑖𝑛𝑘𝜃 3C

𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃) Further complex numbers Proving that: 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 You need to be able to prove that [r(cosθ + isinθ)]n = rn(cos(nθ + isinnθ) for any integer n De Moivre’s theorem can be proved using the method of proof by induction from FP1 Basis – show the statement is true for n = 1 Assumption – assume the statement is true for n = k Inductive – show that if true for n = k, then the statement is also true for n = k + 1 Conclusion – because the statement is true for n = 1 and also true if any value is, then the statement is true for all values of n is true for all positive integers INDUCTIVE  Show that if true for n = k, the statement is also true for n = k + 1 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑛 Sub in n = k + 1 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑘+1 You can write this as two separate parts as the powers are added together 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑘 × 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 1 We can rewrite the first part based on the assumption step, and the second based on the basis step = 𝑟 𝑘 𝑐𝑜𝑠𝑘𝜃+𝑖𝑠𝑖𝑛𝑘𝜃 ×𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) Using the multiplication rules from 3B  Multiply the moduli, add the arguments = 𝑟 𝑘+1 cos⁡(𝑘+1)𝜃+𝑖𝑠𝑖𝑛(𝑘+1)𝜃 3C

𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃) Further complex numbers Proving that: 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 You need to be able to prove that [r(cosθ + isinθ)]n = rn(cos(nθ + isinnθ) for any integer n De Moivre’s theorem can be proved using the method of proof by induction from FP1 Basis – show the statement is true for n = 1 Assumption – assume the statement is true for n = k Inductive – show that if true for n = k, then the statement is also true for n = k + 1 Conclusion – because the statement is true for n = 1 and also true if any value is, then the statement is true for all values of n is true for all positive integers CONCLUSION Explain why this shows it is true… We showed the statement is true for n = 1 We then assumed the following: 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑘 = 𝑟 𝑘 𝑐𝑜𝑠𝑘𝜃+𝑖𝑠𝑖𝑛𝑘𝜃 Using the assumption, we showed that: 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑘+1 = 𝑟 𝑘+1 cos 𝑘+1 𝜃+𝑖𝑠𝑖𝑛(𝑘+1)𝜃 As all the ‘k’ terms have become ‘k + 1’ terms, if the statement is true for one term, it must be true for the next, and so on…  The statement was true for 1, so must be true for 2, and therefore 3, and so on…  We have therefore proven the statement for all positive integers! 3C

𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃) Further complex numbers 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 −𝑚 You need to be able to prove that [r(cosθ + isinθ)]n = rn(cos(nθ + isinnθ) for any integer n We have just proved the theorem for n = k where k is a positive integer Now we need to show it is also true for any negative integer… If n is a negative integer, it can be written as ‘-m’, where m is a positive integer Write using a positive power instead = 1 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑚 Use De Moivre’s theorem for a positive number (which we have proved) = 1 𝑟 𝑚 𝑐𝑜𝑠𝑚𝜃+𝑖𝑠𝑖𝑛𝑚𝜃 Multiply to change some terms in the fraction = 1 𝑟 𝑚 𝑐𝑜𝑠𝑚𝜃+𝑖𝑠𝑖𝑛𝑚𝜃 × 𝑐𝑜𝑠𝑚𝜃−𝑖𝑠𝑖𝑛𝑚𝜃 𝑐𝑜𝑠𝑚𝜃−𝑖𝑠𝑖𝑛𝑚𝜃 Multiply out like quadratics – the bottom is the difference of two squares = 𝑐𝑜𝑠𝑚𝜃−𝑖𝑠𝑖𝑛𝑚𝜃 𝑟 𝑚 𝑐𝑜𝑠 2 𝑚𝜃− 𝑖 2 𝑠𝑖𝑛 2 𝑚𝜃 i2 = -1 = 𝑐𝑜𝑠𝑚𝜃−𝑖𝑠𝑖𝑛𝑚𝜃 𝑟 𝑚 𝑐𝑜𝑠 2 𝑚𝜃+ 𝑠𝑖𝑛 2 𝑚𝜃 You can see that the answer has followed the same pattern as De Moivre’s theorem! You can cancel the denominator as it is equal to 1 = 1 𝑟 𝑚 𝑐𝑜𝑠𝑚𝜃−𝑖𝑠𝑖𝑛𝑚𝜃 Use cos(-θ) = cos(θ) and sin(-θ) = -sinθ = 𝑟 −𝑚 cos −𝑚𝜃 +𝑖𝑠𝑖𝑛(−𝑚𝜃) 3C

𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃) Further complex numbers You need to be able to prove that [r(cosθ + isinθ)]n = rn(cos(nθ + isinnθ) for any integer n Having now proved that De Moivre’s theorem works for both positive and negative integers, there is only one left We need to prove it is true for 0! This is straightforward. As it is just a single value, we can substitute it in to see what happens… 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 Sub in n = 0 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 0 = 𝑟 0 𝑐𝑜𝑠0+𝑖𝑠𝑖𝑛0 Left side = 1 as anything to the power 0 is 1  You can find cos0 and sin 0 as well 1= 1(1+0) ‘Calculate’ 1= 1 So we have shown that De Moivre’s Theorem is true for all positive integers, all negative integers and 0’  It is therefore true for all integers! 3C

𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃) 𝑧 𝑛 = 𝑟𝑒 𝑖𝜃 𝑛 = 𝑟 𝑛 𝑒 𝑖𝑛𝜃 Further complex numbers You need to be able to prove that [r(cosθ + isinθ)]n = rn(cos(nθ + isinnθ) for any integer n It is important to note that De Moivre’s theorem can also be used in exponential form. 𝑟𝑒 𝑖𝜃 𝑛 Both parts will be raised to the power ‘n’ = 𝑟 𝑛 𝑒 𝑖𝑛𝜃 You can remove the bracket! = 𝑟 𝑛 𝑒 𝑖𝑛𝜃 This is De Moivre’s theorem in exponential form! 3C

𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃) 𝑧 𝑛 = 𝑟𝑒 𝑖𝜃 𝑛 = 𝑟 𝑛 𝑒 𝑖𝑛𝜃 Further complex numbers 𝑐𝑜𝑠 9𝜋 17 +𝑖𝑠𝑖𝑛 9𝜋 𝑐𝑜𝑠 2𝜋 17 −𝑖𝑠𝑖𝑛 2𝜋 You need to be able to prove that [r(cosθ + isinθ)]n = rn(cos(nθ + isinnθ) for any integer n Simplify the following: The denominator has to have the ‘+’ sign in the middle  Apply cos(-θ) = cosθ and sin(-θ) = -sinθ 𝑐𝑜𝑠 9𝜋 17 +𝑖𝑠𝑖𝑛 9𝜋 𝑐𝑜𝑠 − 2𝜋 17 +𝑖𝑠𝑖𝑛 − 2𝜋 𝑐𝑜𝑠 9𝜋 17 +𝑖𝑠𝑖𝑛 9𝜋 𝑐𝑜𝑠 2𝜋 17 −𝑖𝑠𝑖𝑛 2𝜋 Apply De Moivre’s theorem (there is no modulus value to worry about here!)  Just multiply the arguments by the power 𝑐𝑜𝑠 45𝜋 17 +𝑖𝑠𝑖𝑛 45𝜋 17 𝑐𝑜𝑠 − 6𝜋 17 +𝑖𝑠𝑖𝑛 − 6𝜋 17 Apply the rules from 3B for the division of complex numbers  Divide the moduli and subtract the arguments 𝑐𝑜𝑠 45𝜋 17 −− 6𝜋 17 +𝑖𝑠𝑖𝑛 45𝜋 17 −− 6𝜋 17 Simplify the sin and cos terms 𝑐𝑜𝑠3𝜋+𝑖𝑠𝑖𝑛3𝜋 Calculate the sin and cos terms =−1+0𝑖 Simplify =−1 3C

𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃) 𝑧 𝑛 = 𝑟𝑒 𝑖𝜃 𝑛 = 𝑟 𝑛 𝑒 𝑖𝑛𝜃 Further complex numbers y You need to be able to prove that [r(cosθ + isinθ)]n = rn(cos(nθ + isinnθ) for any integer n Express the following in the form x + iy where x Є R and y Є R You need to write this in one of the forms above, and you can then use De Moivre’s theorem This is easier than raising a bracket to the power 7! Start with an argand diagram to help find the modulus and argument of the part in the bracket Pay attention to the directions The ‘x’ part is positive so will go in the positive direction horizontally The ‘y’ part is positive so will go upwards r √3 θ x 1 1+ 3 𝑖 7 𝑟= 𝑇𝑎𝑛𝜃= Calculate Inverse Tan 𝑟=2 𝜃= 𝜋 3 𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 Sub in r and θ 2 𝑐𝑜𝑠 𝜋 3 +𝑖𝑠𝑖𝑛 𝜋 3 3C

𝑧 𝑛 = 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 (𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃) 𝑧 𝑛 = 𝑟𝑒 𝑖𝜃 𝑛 = 𝑟 𝑛 𝑒 𝑖𝑛𝜃 Further complex numbers You need to be able to prove that [r(cosθ + isinθ)]n = rn(cos(nθ + isinnθ) for any integer n Express the following in the form x + iy where x Є R and y Є R You need to write this in one of the forms above, and you can then use De Moivre’s theorem This is easier than raising a bracket to the power 7! Start with an argand diagram to help find the modulus and argument of the part in the bracket 1+ 3 𝑖 7 Rewrite using the different form we worked out before 2 𝑐𝑜𝑠 𝜋 3 +𝑖𝑠𝑖𝑛 𝜋 Use De Moivre’s Theorem as above 1+ 3 𝑖 7 2 7 𝑐𝑜𝑠 7𝜋 3 +𝑖𝑠𝑖𝑛 7𝜋 3 Calculate the cos and sin parts = 𝑖 Multiply out and simplify = 𝑖 3C

Teachings for exercise 3D

𝑎+𝑏 𝑛 = 𝑎 𝑛 +𝑛 𝐶 1 𝑎 𝑛−1 𝑏+𝑛 𝐶 2 𝑎 𝑛−2 𝑏 2 +𝑛 𝐶 3 𝑎 𝑛−3 𝑏 3 + ………… + 𝑏 𝑛 Further complex numbers 𝑎+𝑏 𝑛 You can apply De Moivre’s theorem to trigonometric identities This involves changing expressions involving a function of θ into one without. For example changing a cos6θ into powers of cosθ You will need to use the binomial expansion for C2 in this section 𝑎 𝑛 +𝑛 𝐶 1 𝑎 𝑛−1 𝑏+𝑛 𝐶 2 𝑎 𝑛−2 𝑏 2 +𝑛 𝐶 3 𝑎 𝑛−3 𝑏 3 + ………… + 𝑏 𝑛 Remember nCr is a function you can find on your calculator The first term has the full power of n  As you move across you slowly swap the powers over to the second term until it has the full power of n For example: 𝑎+𝑏 4 Follow the pattern above 𝑎 4 + 4 𝐶 1 𝑎 3 𝑏 + 4 𝐶 2 𝑎 2 𝑏 2 + 4 𝐶 3 𝑎 𝑏 3 + 𝑏 4 You can work out the nCr parts 𝑎 4 + 4 𝑎 3 𝑏 + 6 𝑎 2 𝑏 2 + 4𝑎 𝑏 3 + 𝑏 4 3D

𝑎+𝑏 𝑛 = 𝑎 𝑛 +𝑛 𝐶 1 𝑎 𝑛−1 𝑏+𝑛 𝐶 2 𝑎 𝑛−2 𝑏 2 +𝑛 𝐶 3 𝑎 𝑛−3 𝑏 3 + ………… + 𝑏 𝑛 Further complex numbers 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 3 You can apply De Moivre’s theorem to trigonometric identities Express cos3θ using powers of cosθ. This type of question involves making a comparison between two processes One which will give you a ‘cos3θ’ term – you will use De Moivre’s Theorem for this One which will give you an expression in terms of cosθ – you will use the binomial expansion for this You have to think logically and decide where to start If we apply De Moivre’s theorem to this, we will end up with a ‘cos3θ’ term If we apply the binomial expansion to it, we will end up with some terms with cosθ in So this expression is a good starting point! 3D

𝑎+𝑏 𝑛 = 𝑎 𝑛 +𝑛 𝐶 1 𝑎 𝑛−1 𝑏+𝑛 𝐶 2 𝑎 𝑛−2 𝑏 2 +𝑛 𝐶 3 𝑎 𝑛−3 𝑏 3 + ………… + 𝑏 𝑛 Further complex numbers 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 3 You can apply De Moivre’s theorem to trigonometric identities Express cos3θ using powers of cosθ. This type of question involves making a comparison between two processes One which will give you a ‘cos3θ’ term – you will use De Moivre’s Theorem for this One which will give you an expression in terms of cosθ – you will use the binomial expansion for this You have to think logically and decide where to start Apply De Moivre’s theorem 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 3 Follow the rules you know =𝑐𝑜𝑠3𝜃+𝑖𝑠𝑖𝑛3𝜃 Apply the Binomial expansion 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 3 Write out = 𝑐𝑜𝑠𝜃 3 + 3 𝐶 1 𝑐𝑜𝑠𝜃 2 𝑖𝑠𝑖𝑛𝜃 + 3 𝐶 2 𝑐𝑜𝑠𝜃 𝑖𝑠𝑖𝑛𝜃 2 + 𝑖𝑠𝑖𝑛𝜃 3 ‘Tidy up’ = 𝑐𝑜𝑠 3 𝜃 + 3 𝑖𝑐𝑜𝑠 2 𝜃𝑠𝑖𝑛𝜃 + 3 𝑖 2 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 2 𝜃 + 𝑖 3 𝑠𝑖𝑛 3 𝜃 Replace i2 parts with -1 = 𝑐𝑜𝑠 3 𝜃 + 3 𝑖𝑐𝑜𝑠 2 𝜃𝑠𝑖𝑛𝜃 − 3𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 2 𝜃 − 𝑖 𝑠𝑖𝑛 3 𝜃 The two expressions we have made must be equal Therefore the real parts in each and the imaginary parts in each must be the same Equate the real parts 𝑐𝑜𝑠3𝜃= 𝑐𝑜𝑠 3 𝜃−3𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 2 𝜃 3D

𝑎+𝑏 𝑛 = 𝑎 𝑛 +𝑛 𝐶 1 𝑎 𝑛−1 𝑏+𝑛 𝐶 2 𝑎 𝑛−2 𝑏 2 +𝑛 𝐶 3 𝑎 𝑛−3 𝑏 3 + ………… + 𝑏 𝑛 Further complex numbers You can apply De Moivre’s theorem to trigonometric identities Express cos3θ using powers of cosθ. This type of question involves making a comparison between two processes One which will give you a ‘cos3θ’ term – you will use De Moivre’s Theorem for this One which will give you an expression in terms of cosθ – you will use the binomial expansion for this You have to think logically and decide where to start 𝑐𝑜𝑠3𝜃= 𝑐𝑜𝑠 3 𝜃−3𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 2 𝜃 Replace sin2θ with an expression in cos2θ 𝑐𝑜𝑠3𝜃= 𝑐𝑜𝑠 3 𝜃−3𝑐𝑜𝑠𝜃 1− 𝑐𝑜𝑠 2 𝜃 Expand the bracket 𝑐𝑜𝑠3𝜃= 𝑐𝑜𝑠 3 𝜃−3𝑐𝑜𝑠𝜃+3 𝑐𝑜𝑠 3 𝜃 Simplify 𝑐𝑜𝑠3𝜃= 4𝑐𝑜𝑠 3 𝜃−3𝑐𝑜𝑠𝜃 We have successfully expressed cos3θ as posers of cosθ! 3D

𝑎+𝑏 𝑛 = 𝑎 𝑛 +𝑛 𝐶 1 𝑎 𝑛−1 𝑏+𝑛 𝐶 2 𝑎 𝑛−2 𝑏 2 +𝑛 𝐶 3 𝑎 𝑛−3 𝑏 3 + ………… + 𝑏 𝑛 Further complex numbers 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 6 You can apply De Moivre’s theorem to trigonometric identities Express the following as powers of cosθ: So we need something that will give us sin6θ using De Moivre’s theorem It also needs to give us terms of cosθ from the binomial expansion If we apply De Moivre’s theorem to this, we will end up with a ‘sin6θ’ term If we apply the binomial expansion to it, we will end up with some terms with cosθ in So this expression is a good starting point! (and yes you will have to do some expansions larger than powers of 3 or 4!) 𝑠𝑖𝑛6𝜃 𝑠𝑖𝑛𝜃 3D

𝑎+𝑏 𝑛 = 𝑎 𝑛 +𝑛 𝐶 1 𝑎 𝑛−1 𝑏+𝑛 𝐶 2 𝑎 𝑛−2 𝑏 2 +𝑛 𝐶 3 𝑎 𝑛−3 𝑏 3 + ………… + 𝑏 𝑛 Further complex numbers 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 6 You can apply De Moivre’s theorem to trigonometric identities Express the following as powers of cosθ: Apply De Moivre’s theorem 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 6 Follow the rules you know =𝑐𝑜𝑠6𝜃+𝑖𝑠𝑖𝑛6𝜃 𝑠𝑖𝑛6𝜃 𝑠𝑖𝑛𝜃 Apply the Binomial expansion 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 6 = 𝑐𝑜𝑠𝜃 6 + 6 𝐶 1 𝑐𝑜𝑠𝜃 5 (𝑖𝑠𝑖𝑛𝜃) + 6 𝐶 2 𝑐𝑜𝑠𝜃 4 𝑖𝑠𝑖𝑛𝜃 2 + 6 𝐶 3 𝑐𝑜𝑠𝜃 3 𝑖𝑠𝑖𝑛𝜃 3 + 6 𝐶 4 𝑐𝑜𝑠𝜃 2 𝑖𝑠𝑖𝑛𝜃 4 + 6 𝐶 5 𝑐𝑜𝑠𝜃 𝑖𝑠𝑖𝑛𝜃 5 + 𝑖𝑠𝑖𝑛𝜃 6 = 𝑐𝑜𝑠 6 𝜃 + 6𝑖 𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃 + 15 𝑖 2 𝑐𝑜𝑠 4 𝜃 𝑠𝑖𝑛 2 𝜃 + 20 𝑖 3 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛 3 𝜃 + 15 𝑖 4 𝑐𝑜𝑠 2 𝜃 𝑠𝑖𝑛 4 𝜃 + 6 𝑖 5 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 5 𝜃 + 𝑖 6 𝑠𝑖𝑛 6 𝜃 Replace terms: i2 and i6 = -1 i4 = 1 = 𝑐𝑜𝑠 6 𝜃 + 6𝑖 𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃 − 15 𝑐𝑜𝑠 4 𝜃 𝑠𝑖𝑛 2 𝜃 − 20𝑖 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛 3 𝜃 + 15 𝑐𝑜𝑠 2 𝜃 𝑠𝑖𝑛 4 𝜃 + 6𝑖𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 5 𝜃 − 𝑠𝑖𝑛 6 𝜃 So the two expressions created from De Moivre and the Binomial Expansion must be equal The real parts will be the same, as will the imaginary parts This time we have to equate the imaginary parts as this has sin6θ in 𝑖𝑠𝑖𝑛6𝜃=6𝑖 𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃−20𝑖 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛 3 𝜃+6𝑖𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 5 𝜃 Divide all by i 𝑠𝑖𝑛6𝜃=6 𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃−20 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛 3 𝜃+6𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 5 𝜃 3D

𝑎+𝑏 𝑛 = 𝑎 𝑛 +𝑛 𝐶 1 𝑎 𝑛−1 𝑏+𝑛 𝐶 2 𝑎 𝑛−2 𝑏 2 +𝑛 𝐶 3 𝑎 𝑛−3 𝑏 3 + ………… + 𝑏 𝑛 Further complex numbers 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 6 You can apply De Moivre’s theorem to trigonometric identities Express the following as powers of cosθ: 𝑠𝑖𝑛6𝜃=6 𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃−20 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛 3 𝜃+6𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 5 𝜃 So we have changed the expression we were given into powers of cosθ! 𝑠𝑖𝑛6𝜃 𝑠𝑖𝑛𝜃 2 4 𝑠𝑖𝑛6𝜃 𝑠𝑖𝑛𝜃 = 6 𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃−20 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛 3 𝜃+6𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 5 𝜃 𝑠𝑖𝑛𝜃 Divide all terms by sinθ =6 𝑐𝑜𝑠 5 𝜃−20 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛 2 𝜃+6𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 4 𝜃 Replace sin2θ terms with (1 – cos2θ) terms =6 𝑐𝑜𝑠 5 𝜃−20 𝑐𝑜𝑠 3 𝜃 1−𝑐𝑜𝑠 2 𝜃 +6𝑐𝑜𝑠𝜃 1−𝑐𝑜𝑠 2 𝜃 2 Expand the first bracket, square the second =6 𝑐𝑜𝑠 5 𝜃 − 20 𝑐𝑜𝑠 3 𝜃 + 20 𝑐𝑜𝑠 5 𝜃 + 6𝑐𝑜𝑠𝜃 1− 2𝑐𝑜𝑠 2 𝜃+ 𝑐𝑜𝑠 4 𝜃 Expand the second bracket =6 𝑐𝑜𝑠 5 𝜃 − 20 𝑐𝑜𝑠 3 𝜃 + 20 𝑐𝑜𝑠 5 𝜃 + 6𝑐𝑜𝑠𝜃 − 12 𝑐𝑜𝑠 3 𝜃 + 6 𝑐𝑜𝑠 5 𝜃 Group together the like terms =32 𝑐𝑜𝑠 5 𝜃 − 32 𝑐𝑜𝑠 3 𝜃 + 6𝑐𝑜𝑠𝜃 3D

𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃 Further complex numbers Let: 𝑧=𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 You can apply De Moivre’s theorem to trigonometric identities You also need to be able to work this type of question in a different way: For example, you might have a power or cos or sin and need to express it using several linear terms instead Eg) Changing sin6θ to sinaθ + sinbθ where a and b are integers  To do this we need to know some other patterns first! Write as ‘1 over’  or with a power of -1 1 𝑧 = 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 −1 Use De Moivre’s theorem 1 𝑧 =cos⁡(−𝜃)+𝑖𝑠𝑖𝑛(−𝜃) Use cos(-θ) = cosθ and sin(-θ) = -sinθ 1 𝑧 = cos 𝜃 −𝑖𝑠𝑖𝑛𝜃 We can add our two results together: 𝑧+ 1 𝑧 = 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 + 𝑐𝑜𝑠𝜃−𝑖𝑠𝑖𝑛𝜃 Simplify 𝑧+ 1 𝑧 = 2𝑐𝑜𝑠𝜃 3D

𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃 𝑧− 1 𝑧 =2𝑖𝑠𝑖𝑛𝜃 Further complex numbers Let: 𝑧=𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 You can apply De Moivre’s theorem to trigonometric identities You also need to be able to work this type of question in a different way: For example, you might have a power or cos or sin and need to express it using several linear terms instead Eg) Changing sin6θ to sinaθ + sinbθ where a and b are integers  To do this we need to know some other patterns first! Write as ‘1 over’  or with a power of -1 1 𝑧 = 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 −1 Use De Moivre’s theorem 1 𝑧 =cos⁡(−𝜃)+𝑖𝑠𝑖𝑛(−𝜃) Use cos(-θ) = cosθ and sin(-θ) = -sinθ 1 𝑧 = cos 𝜃 −𝑖𝑠𝑖𝑛𝜃 We could also subtract our two results: 𝑧− 1 𝑧 = 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 − 𝑐𝑜𝑠𝜃−𝑖𝑠𝑖𝑛𝜃 Simplify 𝑧− 1 𝑧 = 2𝑖𝑠𝑖𝑛𝜃 3D

𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃 𝑧− 1 𝑧 =2𝑖𝑠𝑖𝑛𝜃 𝑧 𝑛 + 1 𝑧 𝑛 =2𝑐𝑜𝑠𝑛𝜃 Further complex numbers Let: 𝑧 𝑛 =𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 You can apply De Moivre’s theorem to trigonometric identities You also need to be able to work this type of question in a different way: For example, you might have a power or cos or sin and need to express it using several linear terms instead Eg) Changing sin6θ to sinaθ + sinbθ where a and b are integers To do this we need to know some other patterns first! You can also apply the rules we just saw to powers of z Write as ‘1 over’  or with a power of -1 1 𝑧 𝑛 = 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 −1 Use De Moivre’s theorem 1 𝑧 𝑛 =cos⁡(−𝑛𝜃)+𝑖𝑠𝑖𝑛(−𝑛𝜃) Use cos(-θ) = cosθ and sin(-θ) = -sinθ 1 𝑧 𝑛 = cos 𝑛𝜃 −𝑖𝑠𝑖𝑛𝑛𝜃 We could add our two results together: 𝑧 𝑛 + 1 𝑧 𝑛 = 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 + 𝑐𝑜𝑠𝑛𝜃−𝑖𝑠𝑖𝑛𝑛𝜃 Simplify 𝑧 𝑛 + 1 𝑧 𝑛 = 2𝑐𝑜𝑠𝑛𝜃 3D

𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃 𝑧− 1 𝑧 =2𝑖𝑠𝑖𝑛𝜃 𝑧 𝑛 + 1 𝑧 𝑛 =2𝑐𝑜𝑠𝑛𝜃 𝑧 𝑛 − 1 𝑧 𝑛 =2𝑖𝑠𝑖𝑛𝑛𝜃 Further complex numbers Let: 𝑧 𝑛 =𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 You can apply De Moivre’s theorem to trigonometric identities You also need to be able to work this type of question in a different way: For example, you might have a power or cos or sin and need to express it using several linear terms instead Eg) Changing sin6θ to sinaθ + sinbθ where a and b are integers To do this we need to know some other patterns first! You can also apply the rules we just saw to powers of z Write as ‘1 over’  or with a power of -1 1 𝑧 𝑛 = 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 −1 Use De Moivre’s theorem 1 𝑧 𝑛 =cos⁡(−𝑛𝜃)+𝑖𝑠𝑖𝑛(−𝑛𝜃) Use cos(-θ) = cosθ and sin(-θ) = -sinθ 1 𝑧 𝑛 = cos 𝑛𝜃 −𝑖𝑠𝑖𝑛𝑛𝜃 We could also subtract our two results: 𝑧 𝑛 − 1 𝑧 𝑛 = 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 − 𝑐𝑜𝑠𝑛𝜃−𝑖𝑠𝑖𝑛𝑛𝜃 Simplify 𝑧 𝑛 − 1 𝑧 𝑛 = 2𝑖𝑠𝑖𝑛𝑛𝜃 3D

𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃 𝑧− 1 𝑧 =2𝑖𝑠𝑖𝑛𝜃 𝑧 𝑛 + 1 𝑧 𝑛 =2𝑐𝑜𝑠𝑛𝜃 𝑧 𝑛 − 1 𝑧 𝑛 =2𝑖𝑠𝑖𝑛𝑛𝜃 Further complex numbers Using the Identity above Creating a cos5θ term You can apply De Moivre’s theorem to trigonometric identities Let’s now see how we can use these ‘patterns’ in solving problems: Express cos5θ in the form acos5θ + bcos3θ + ccosθ Where a, b and c are constants to be found. When working this way round you need to use the identities above to express both cos5θ and terms with cos5θ, cos3θ and cosθ. You can then set the expressions equal to each other 𝑧+ 1 𝑧 5 = 2𝑐𝑜𝑠𝜃 5 =32 𝑐𝑜𝑠 5 𝜃 Creating the other cos terms – use the Binomial expansion! 𝑧+ 1 𝑧 5 Use the B.E. + 5 𝑧 𝑧 + 10 𝑧 𝑧 2 + 10 𝑧 𝑧 3 + 5𝑧 1 𝑧 4 𝑧 5 = 𝑧 5 Cancel some z terms 𝑧 𝑧 3 𝑧 5 = 𝑧 5 + 5𝑧 3 + 10𝑧 Group up terms with the same power = 𝑧 𝑧 5 + 5 𝑧 𝑧 3 + 10 𝑧+ 1 𝑧 Rewrite using an identity above =2𝑐𝑜𝑠5𝜃 + 5(2𝑐𝑜𝑠3𝜃) + 10(2𝑐𝑜𝑠𝜃) Simplify =2𝑐𝑜𝑠5𝜃 + 10𝑐𝑜𝑠3𝜃 + 20𝑐𝑜𝑠𝜃 3D

𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃 𝑧− 1 𝑧 =2𝑖𝑠𝑖𝑛𝜃 𝑧 𝑛 + 1 𝑧 𝑛 =2𝑐𝑜𝑠𝑛𝜃 𝑧 𝑛 − 1 𝑧 𝑛 =2𝑖𝑠𝑖𝑛𝑛𝜃 Further complex numbers Using the two expressions You can apply De Moivre’s theorem to trigonometric identities Let’s now see how we can use these ‘patterns’ in solving problems: Express cos5θ in the form acos5θ + bcos3θ + ccosθ Where a, b and c are constants to be found. When working this way round you need to use the identities above to express both cos5θ and terms with cos5θ, cos3θ and cosθ. You can then set the expressions equal to each other 𝑧+ 1 𝑧 5 These two expressions must be equal to each other =32 𝑐𝑜𝑠 5 𝜃 𝑧+ 1 𝑧 5 =2𝑐𝑜𝑠5𝜃+10𝑐𝑜𝑠3𝜃+ 20𝑐𝑜𝑠𝜃 32 𝑐𝑜𝑠 5 𝜃=2𝑐𝑜𝑠5𝜃+10𝑐𝑜𝑠3𝜃+ 20𝑐𝑜𝑠𝜃 Divide both sides by 32 𝑐𝑜𝑠 5 𝜃= 1 16 𝑐𝑜𝑠5𝜃 𝑐𝑜𝑠3𝜃 𝑐𝑜𝑠𝜃 So we have written cos5θ using cos5θ, cos3θ and cosθ 3D

𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃 𝑧− 1 𝑧 =2𝑖𝑠𝑖𝑛𝜃 𝑧 𝑛 + 1 𝑧 𝑛 =2𝑐𝑜𝑠𝑛𝜃 𝑧 𝑛 − 1 𝑧 𝑛 =2𝑖𝑠𝑖𝑛𝑛𝜃 Further complex numbers Using the Identity above Creating a sin3θ term You can apply De Moivre’s theorem to trigonometric identities Show that:  This is similar to the previous question. You need to use rules above to find a way to create a sin3θ expression, and an expression containing sin3θ and sinθ 𝑧− 1 𝑧 3 = 2𝑖𝑠𝑖𝑛𝜃 3 =8 𝑖 3 𝑠𝑖𝑛 3 𝜃 =−8𝑖 𝑠𝑖𝑛 3 𝜃 𝑠𝑖𝑛 3 𝜃=− 1 4 𝑠𝑖𝑛3𝜃+ 3 4 𝑠𝑖𝑛𝜃 Creating the other sin terms – use the Binomial expansion! 𝑧− 1 𝑧 3 Use the B.E. + 3 𝑧 2 − 1 𝑧 + 3𝑧 − 1 𝑧 2 + − 1 𝑧 3 = 𝑧 3 Cancel some z terms 𝑧 − 1 𝑧 3 = 𝑧 3 − 3𝑧 Group up terms with the same power = 𝑧 3 − 1 𝑧 3 − 3 𝑧− 1 𝑧 Rewrite using an identity above =2𝑖𝑠𝑖𝑛3𝜃 − 3(2𝑖𝑠𝑖𝑛𝜃) Simplify =2𝑖𝑠𝑖𝑛3𝜃 − 6𝑖𝑠𝑖𝑛𝜃 3D

𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃 𝑧− 1 𝑧 =2𝑖𝑠𝑖𝑛𝜃 𝑧 𝑛 + 1 𝑧 𝑛 =2𝑐𝑜𝑠𝑛𝜃 𝑧 𝑛 − 1 𝑧 𝑛 =2𝑖𝑠𝑖𝑛𝑛𝜃 Further complex numbers Using the two expressions You can apply De Moivre’s theorem to trigonometric identities Show that:  This is similar to the previous question. You need to use rules above to find a way to create a sin3θ expression, and an expression containing sin3θ and sinθ 𝑧− 1 𝑧 3 These two expressions must be equal to each other =−8𝑖 𝑠𝑖𝑛 3 𝜃 𝑧− 1 𝑧 3 𝑠𝑖𝑛 3 𝜃=− 1 4 𝑠𝑖𝑛3𝜃+ 3 4 𝑠𝑖𝑛𝜃 =2𝑖𝑠𝑖𝑛3𝜃−6𝑖𝑠𝑖𝑛𝜃 −8𝑖 𝑠𝑖𝑛 3 𝜃=2𝑖𝑠𝑖𝑛3𝜃−6𝑖𝑠𝑖𝑛𝜃 Divide both sides by i −8 𝑠𝑖𝑛 3 𝜃=2𝑠𝑖𝑛3𝜃−6𝑠𝑖𝑛𝜃 Divide both sides by -8 𝑠𝑖𝑛 3 𝜃=− 1 4 𝑠𝑖𝑛3𝜃+ 3 4 𝑠𝑖𝑛𝜃 So we have written sin3θ using sin3θ and sinθ! 3D

𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃 𝑧− 1 𝑧 =2𝑖𝑠𝑖𝑛𝜃 𝑧 𝑛 + 1 𝑧 𝑛 =2𝑐𝑜𝑠𝑛𝜃 𝑧 𝑛 − 1 𝑧 𝑛 =2𝑖𝑠𝑖𝑛𝑛𝜃 Further complex numbers Using the Identity above Creating a sin4θ term You can apply De Moivre’s theorem to trigonometric identities a) Express sin4θ in the form: Where d, e and f are constants to be found. b) Hence, find the exact value of the following integral:  Start exactly as with the previous questions, by finding an expression with sin4θ and one with cos4θ, cos2θ and a number 𝑧− 1 𝑧 4 = 2𝑖𝑠𝑖𝑛𝜃 4 =16 𝑖 4 𝑠𝑖𝑛 4 𝜃 =16 𝑠𝑖𝑛 4 𝜃 𝑑𝑐𝑜𝑠4𝜃+𝑒𝑐𝑜𝑠2𝜃+𝑓 Creating the cos terms – use the Binomial expansion! 𝑧− 1 𝑧 4 Use the B.E. + 4 𝑧 3 − 1 𝑧 + 6 𝑧 2 − 1 𝑧 2 + 4𝑧 − 1 𝑧 3 + − 1 𝑧 4 = 𝑧 4 Cancel some z terms − 𝑧 2 𝑧 4 = 𝑧 4 − 4 𝑧 2 + 6 0 𝜋 2 𝑠𝑖𝑛 4 𝜃 𝑑𝜃 Group up terms with the same power (use positive values in the brackets so we get cos terms) = 𝑧 𝑧 4 − 4 𝑧 𝑧 2 + 6 Replace using an identity above =2𝑐𝑜𝑠4𝜃 − 4(2𝑐𝑜𝑠2𝜃) + 6 Simplify =2𝑐𝑜𝑠4𝜃 − 8𝑐𝑜𝑠2𝜃 + 6 3D

𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃 𝑧− 1 𝑧 =2𝑖𝑠𝑖𝑛𝜃 𝑧 𝑛 + 1 𝑧 𝑛 =2𝑐𝑜𝑠𝑛𝜃 𝑧 𝑛 − 1 𝑧 𝑛 =2𝑖𝑠𝑖𝑛𝑛𝜃 Further complex numbers Using the two expressions You can apply De Moivre’s theorem to trigonometric identities a) Express sin4θ in the form: Where d, e and f are constants to be found. b) Hence, find the exact value of the following integral:  Start exactly as with the previous questions, by finding an expression with sin4θ and one with cos4θ, cos2θ and a number 𝑧− 1 𝑧 4 These two expressions must be equal to each other =16 𝑠𝑖𝑛 4 𝜃 𝑧− 1 𝑧 4 𝑑𝑐𝑜𝑠4𝜃+𝑒𝑐𝑜𝑠2𝜃+𝑓 =2𝑐𝑜𝑠4𝜃−8𝑐𝑜𝑠2𝜃+6 16 𝑠𝑖𝑛 4 𝜃=2𝑐𝑜𝑠4𝜃−8𝑐𝑜𝑠2𝜃+6 Divide both sides by 16 𝑠𝑖𝑛 4 𝜃= 1 8 𝑐𝑜𝑠4𝜃− 1 2 𝑐𝑜𝑠2𝜃+ 3 8 0 𝜋 2 𝑠𝑖𝑛 4 𝜃 𝑑𝜃 So we have written sin4θ using cos4θ and cos2θ! 3D

𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃 𝑧− 1 𝑧 =2𝑖𝑠𝑖𝑛𝜃 𝑧 𝑛 + 1 𝑧 𝑛 =2𝑐𝑜𝑠𝑛𝜃 𝑧 𝑛 − 1 𝑧 𝑛 =2𝑖𝑠𝑖𝑛𝑛𝜃 Further complex numbers Cosine Integrals (in C4) 𝑠𝑖𝑛 4 𝜃= 1 8 𝑐𝑜𝑠4𝜃− 1 2 𝑐𝑜𝑠2𝜃+ 3 8 𝑐𝑜𝑠4𝜃 = 1 4 𝑠𝑖𝑛4𝜃 You can apply De Moivre’s theorem to trigonometric identities a) Express sin4θ in the form: Where d, e and f are constants to be found. b) Hence, find the exact value of the following integral:  Start exactly as with the previous questions, by finding an expression with sin4θ and one with cos4θ, cos2θ and a number 𝑐𝑜𝑠2𝜃 = 1 2 𝑠𝑖𝑛2𝜃 0 𝜋 2 𝑠𝑖𝑛 4 𝜃 𝑑𝜃 𝑑𝑐𝑜𝑠4𝜃+𝑒𝑐𝑜𝑠2𝜃+𝑓 Replace with an equivalent expression 0 𝜋 𝑐𝑜𝑠4𝜃− 1 2 𝑐𝑜𝑠2𝜃 𝑑𝜃 Integrate each term with respect to θ, using knowledge from C4 = 𝜋 2 1 32 𝑠𝑖𝑛4𝜃 − 1 4 𝑠𝑖𝑛2𝜃 𝜃 0 𝜋 2 𝑠𝑖𝑛 4 𝜃 𝑑𝜃 Sub in limits = 𝑠𝑖𝑛4 𝜋 2 − 1 4 𝑠𝑖𝑛2 𝜋 𝜋 2 − 𝑠𝑖𝑛4 0 − 1 4 𝑠𝑖𝑛 Work out = 3 16 𝜋 3D

Teachings for exercise 3E

𝐼𝑓: 𝑧=𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑇ℎ𝑒𝑛: 𝑧=𝑟 cos⁡(𝜃+2𝑘𝜋)+𝑖𝑠𝑖𝑛(𝜃+2𝑘𝜋) Further complex numbers 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 You can use De Moivre’s theorem to find the nth roots of a complex number You already know how to find real roots of a number, but now we need to find both real roots and imaginary roots! We need to apply the following results: If: Then: where k is an integer This is because we can add multiples of 2π to the argument as it will end up in the same place (2π = 360º) 2) De Moivre’s theorem 𝑧=𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑧=𝑟 cos⁡(𝜃+2𝑘𝜋)+𝑖𝑠𝑖𝑛(𝜃+2𝑘𝜋) 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 3E

𝐼𝑓: 𝑧=𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑇ℎ𝑒𝑛: 𝑧=𝑟 cos⁡(𝜃+2𝑘𝜋)+𝑖𝑠𝑖𝑛(𝜃+2𝑘𝜋) Further complex numbers 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 y In this case the modulus and argument are simple to find! You can use De Moivre’s theorem to find the nth roots of a complex number Solve the equation z3 = 1 and represent your solutions on an Argand diagram. First you need to express z in the modulus-argument form. Use an Argand diagram. Now we know r and θ we can set z3 equal to this expression, when written in the modulus-argument form We can then find an expression for z in terms of k We can then solve this to find the roots of the equation above 𝑟=1 1 x 𝜃=0 𝑧 3 =1 𝑐𝑜𝑠0+𝑖𝑠𝑖𝑛0 Apply the rule above 𝑧 3 = cos⁡(0+2𝑘𝜋)+𝑖𝑠𝑖𝑛(0+2𝑘𝜋) Cube root (use a relevant power) 𝑧= cos 0+2𝑘𝜋 +𝑖𝑠𝑖𝑛(0+2𝑘𝜋) 1 3 Apply De Moivre’s theorem 𝑧= cos 0+2𝑘𝜋 3 +𝑖𝑠𝑖𝑛 0+2𝑘𝜋 3 3E

𝐼𝑓: 𝑧=𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑇ℎ𝑒𝑛: 𝑧=𝑟 cos⁡(𝜃+2𝑘𝜋)+𝑖𝑠𝑖𝑛(𝜃+2𝑘𝜋) Further complex numbers 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 𝑧= 𝑐𝑜𝑠 0+2𝑘𝜋 3 +𝑖𝑠𝑖𝑛 0+2𝑘𝜋 3 You can use De Moivre’s theorem to find the nth roots of a complex number Solve the equation z3 = 1 and represent your solutions on an Argand diagram. We now just need to choose different values for k until we have found all the roots The values of k you choose should keep the argument within the range: -π < θ ≤ π k = 0 𝑧= 𝑐𝑜𝑠 0 +𝑖𝑠𝑖𝑛 0 Sub k = 0 in and calculate the cosine and sine parts 𝑧=1 k = 1 𝑧= 𝑐𝑜𝑠 2𝜋 3 +𝑖𝑠𝑖𝑛 2𝜋 3 Sub k = 1 in and calculate the cosine and sine parts 𝑧=− 1 2 +𝑖 So the roots of z3 = 1 are: k = -1 Sub k = -1 in and calculate the cosine and sine parts  (k = 2 would cause the argument to be outside the range) 𝑧=1, − 1 2 +𝑖 and− 1 2 −𝑖 𝑧= 𝑐𝑜𝑠 − 2𝜋 3 +𝑖𝑠𝑖𝑛 − 2𝜋 3 𝑧=− 1 2 −𝑖 3E

𝐼𝑓: 𝑧=𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑇ℎ𝑒𝑛: 𝑧=𝑟 cos⁡(𝜃+2𝑘𝜋)+𝑖𝑠𝑖𝑛(𝜃+2𝑘𝜋) Further complex numbers 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 y You can use De Moivre’s theorem to find the nth roots of a complex number Solve the equation z3 = 1 and represent your solutions on an Argand diagram. We now just need to choose different values for k until we have found all the roots The values of k you choose should keep the argument within the range: -π < θ ≤ π − 𝟏 𝟐 +𝒊 𝟑 𝟐 2 3 𝜋 𝟏 2 3 𝜋 x 2 3 𝜋 − 𝟏 𝟐 −𝒊 𝟑 𝟐 So the roots of z3 = 1 are: The solutions will all the same distance from the origin The angles between them will also be the same The sum of the roots is always equal to 0 𝑧=1, − 1 2 +𝑖 and− 1 2 −𝑖 3E

𝐼𝑓: 𝑧=𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑇ℎ𝑒𝑛: 𝑧=𝑟 cos⁡(𝜃+2𝑘𝜋)+𝑖𝑠𝑖𝑛(𝜃+2𝑘𝜋) Further complex numbers 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 y Find the modulus and argument You can use De Moivre’s theorem to find the nth roots of a complex number Solve the equation z4 - 2√3i = 2 Give your answers in both the modulus-argument and exponential forms. By rearranging… z4 = 2 + 2√3i As before, use an argand diagram to express the equation in the modulus-argument form Then choose values of k until you have found all the solutions 𝑟= r 𝑟=4 2√3 θ 𝜃= 𝑡𝑎𝑛 − x 𝜃= 𝜋 3 2 𝑧 4 =4 𝑐𝑜𝑠 𝜋 3 +𝑖𝑠𝑖𝑛 𝜋 3 Apply the rule above 𝑧 4 =4 𝑐𝑜𝑠 𝜋 3 +2𝑘𝜋 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝑘𝜋 Take the 4th root of each side 𝑧= 4 𝑐𝑜𝑠 𝜋 3 +2𝑘𝜋 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝑘𝜋 De Moivre’s Theorem 𝑧= 𝑐𝑜𝑠 𝜋 3 +2𝑘𝜋 4 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝑘𝜋 4 Work out the power at the front 𝑧= 2 𝑐𝑜𝑠 𝜋 3 +2𝑘𝜋 4 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝑘𝜋 4 3E

𝐼𝑓: 𝑧=𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑇ℎ𝑒𝑛: 𝑧=𝑟 cos⁡(𝜃+2𝑘𝜋)+𝑖𝑠𝑖𝑛(𝜃+2𝑘𝜋) Further complex numbers 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 𝑧= 2 𝑐𝑜𝑠 𝜋 𝑖𝑠𝑖𝑛 𝜋 3 4 You can use De Moivre’s theorem to find the nth roots of a complex number Solve the equation z4 - 2√3i = 2 Give your answers in both the modulus-argument and exponential forms. By rearranging… z4 = 2 + 2√3i As before, use an argand diagram to express the equation in the modulus-argument form Then choose values of k until you have found all the solutions Sub k = 0 in and simplify (you can leave in this form) k = 0 𝑧= 2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12 𝑧= 2 𝑐𝑜𝑠 𝜋 3 +2𝜋 4 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝜋 4 k = 1 𝑧= 2 𝑐𝑜𝑠 7𝜋 12 +𝑖𝑠𝑖𝑛 7𝜋 12 Choose values of k that keep the argument between –π and π 𝑧= 2 𝑐𝑜𝑠 𝜋 3 −2𝜋 4 +𝑖𝑠𝑖𝑛 𝜋 3 −2𝜋 4 k = -1 𝑧= 2 𝑐𝑜𝑠 − 5𝜋 12 +𝑖𝑠𝑖𝑛 − 5𝜋 12 𝑧= 2 𝑐𝑜𝑠 𝜋 3 −2𝜋 4 +𝑖𝑠𝑖𝑛 𝜋 3 −2𝜋 4 k = -2 𝑧= 2 𝑐𝑜𝑠 𝜋 3 +2𝑘𝜋 4 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝑘𝜋 4 𝑧= 2 𝑐𝑜𝑠 − 11𝜋 12 +𝑖𝑠𝑖𝑛 − 11𝜋 12 3E

𝐼𝑓: 𝑧=𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑇ℎ𝑒𝑛: 𝑧=𝑟 cos⁡(𝜃+2𝑘𝜋)+𝑖𝑠𝑖𝑛(𝜃+2𝑘𝜋) Further complex numbers 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 Solutions in the modulus-argument form You can use De Moivre’s theorem to find the nth roots of a complex number Solve the equation z4 - 2√3i = 2 Give your answers in both the modulus-argument and exponential forms. By rearranging… z4 = 2 + 2√3i As before, use an argand diagram to express the equation in the modulus-argument form Then choose values of k until you have found all the solutions 𝑧= 2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12 𝑧= 2 𝑐𝑜𝑠 − 5𝜋 12 +𝑖𝑠𝑖𝑛 − 5𝜋 12 𝑧= 2 𝑐𝑜𝑠 7𝜋 12 +𝑖𝑠𝑖𝑛 7𝜋 12 𝑧= 2 𝑐𝑜𝑠 − 11𝜋 12 +𝑖𝑠𝑖𝑛 − 11𝜋 12 Solutions in the exponential form 𝑧= 2 𝑒 𝜋 12 𝑖 𝑧= 2 𝑒 − 5𝜋 12 𝑖 𝑧= 2 𝑒 7𝜋 12 𝑖 𝑧= 2 𝑒 − 11𝜋 12 𝑖 𝑧= 2 𝑐𝑜𝑠 𝜋 3 +2𝑘𝜋 4 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝑘𝜋 4 3E

Teachings for exercise 3F

You can use complex numbers to represent a locus of points on an Argand diagram A locus a set of points which obey a rule  You will need to be able to understand Loci based on Argand diagrams x The locus of points a given distance from a point O is a circle x O x A The locus of points equidistant from two fixed points A and B is the perpendicular bisector of line AB B 3F

Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram 𝑧− 𝑧 1 Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram z = x + iy represents a variable point P(x,y) on an Argand diagram z1 = x1 + iy1 represents a fixed point A(x1,y1) on an Argand diagram What is represented by:  It represents the distance between the fixed point A(x1,y1) and the variable point P(x,y) P(x,y) z - z1 z A(x1,y1) z1 𝑧− 𝑧 1 x If we want to get from the fixed point A to the variable point P, we need to travel back along z1 and then out along z (-z1 + z) This can be written as a vector, z – z1 So |z – z1| represents the distance between the fixed point and the variable point! 3F

Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram 𝑧− 𝑧 1 Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram If: Sketch the locus of P(x,y) which is represented by z on an Argand diagram P(x,y) A(5,3) 𝑧−5−3𝑖 =3 x 𝑧−5−3𝑖 =3 𝑧−(5+3𝑖) =3 So we want the locus where the distance between the variable point z and the fixed point (5,3) is equal to 3 Leave z as it is – this is the variable point Put this part in a bracket - This is the fixed point This will be a circle of radius 3 units, centre (5,3) 3F

Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram 𝑧− 𝑧 1 Further complex numbers 𝐼𝑓: 𝑧=𝑥+𝑖𝑦 𝑇ℎ𝑒𝑛: 𝑧 = 𝑥+𝑖𝑦 = 𝑥 2 + 𝑦 2 You can use complex numbers to represent a locus of points on an Argand diagram If: Use an algebraic method to find a Cartesian equation of the locus of z So you have to do this without using the graph you drew We will quickly remind ourselves of something that will be useful for this! y P(x,y) |z| 𝑧−5−3𝑖 =3 y x x If: 𝑧=𝑥+𝑖𝑦 (By Pythagoras’ Theorem) 𝑧 = 𝑥+𝑖𝑦 = 𝑥 2 + 𝑦 2 Then: 3F

Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram 𝑧− 𝑧 1 Further complex numbers 𝐼𝑓: 𝑧=𝑥+𝑖𝑦 𝑇ℎ𝑒𝑛: 𝑧 = 𝑥+𝑖𝑦 = 𝑥 2 + 𝑦 2 You can use complex numbers to represent a locus of points on an Argand diagram If: Use an algebraic method to find a Cartesian equation of the locus of z Now we can find the equation of the locus algebraically… 𝑧−5−3𝑖 =3 Replace z with ‘x + iy’ 𝑥+𝑖𝑦−5−3𝑖 =3 𝑧−5−3𝑖 =3 Group the real and imaginary terms 𝑥−5 +𝑖(𝑦−3) =3 Use the rule above to remove the modulus 𝑥− 𝑦−3 2 =3 Square both sides (𝑥−5) 2 + 𝑦−3 2 =9 You (hopefully) recognise that this is the equation of a circle, radius 3 and with centre (5,3)! 3F

Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram 𝑧− 𝑧 1 Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram As a general rule, the locus of: is a circle of radius r and centre (x1,y1) where z1 = x1 + iy1 𝑧− 𝑧 1 =𝑟 3F

Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram 𝑧− 𝑧 1 Further complex numbers 𝑧− 𝑧 1 =𝑟  Circle radius r and centre (x1,y1) You can use complex numbers to represent a locus of points on an Argand diagram Give a geometrical interpretation of each of the following loci of z: a) 𝑧−3𝑖 =4 d) 2−5𝑖−𝑧 =3 ‘Factorise’ the part inside the modulus  Circle, centre (0,3) radius 4 (−1)(−2+5𝑖+𝑧) =3 You can write this as 2 moduli multiplied −1 𝑧−2+5𝑖 =3 |-1| = 1, put the ‘fixed’ part in a bracket b) 𝑧−(2+3𝑖) =5 𝑧−(2−5𝑖) =3  Circle, centre (2,3) radius 5  Circle, centre (2,-5) radius 3 Effectively for d), you just swap the signs of everything in the modulus, its value will not change  |10 - 8| = | | c) 𝑧+3−5𝑖 =2 Put the ‘fixed’ part in a bracket 𝑧−(−3+5𝑖) =2  Circle, centre (-3,5) radius 2 3F

Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram 𝑧− 𝑧 1 Further complex numbers This is the distance of P(x,y) from the origin (0,0) 𝑧 You can use complex numbers to represent a locus of points on an Argand diagram Sketch the locus of P(x,y) which is represented by z on an Argand diagram, if: We therefore need the set of points that are the same distance from (0,0) and (0,6) This will be the bisector of the line joining the two co-ordinates You can see that it is the line with equation y = 3 𝑧−6𝑖 This is the distance of P(x,y) from (0,6) 𝑧−(6𝑖) y 𝑧 = 𝑧−6𝑖 (0,6) y = 3 x (0,0) 3F

Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram 𝑧− 𝑧 1 Further complex numbers 𝐼𝑓: 𝑧=𝑥+𝑖𝑦 𝑇ℎ𝑒𝑛: 𝑧 = 𝑥+𝑖𝑦 = 𝑥 2 + 𝑦 2 You can use complex numbers to represent a locus of points on an Argand diagram Sketch the locus of P(x,y) which is represented by z on an Argand diagram, if: Show that the locus is y = 3 using an algebraic method 𝑧 = 𝑧−6𝑖 Replace z with x + iy 𝑥+𝑖𝑦 = 𝑥+𝑖𝑦−6𝑖 Factorise the ‘i’ terms on the right side 𝑥+𝑖𝑦 = 𝑥+𝑖(𝑦−6) Use the rule above to remove the moduli 𝑧 = 𝑧−6𝑖 𝑥 2 + 𝑦 2 = 𝑥 2 + 𝑦−6 2 Square both sides 𝑥 2 + 𝑦 2 = 𝑥 2 + 𝑦−6 2 Expand the bracket 𝑥 2 + 𝑦 2 = 𝑥 2 + 𝑦 2 −12𝑦+36 Cancel terms on each side 0=−12𝑦+36 Add 12y 12𝑦=36 Divide by 12 𝑦=3 3F

Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram 𝑧− 𝑧 1 Further complex numbers 𝐼𝑓: 𝑧=𝑥+𝑖𝑦 𝑇ℎ𝑒𝑛: 𝑧 = 𝑥+𝑖𝑦 = 𝑥 2 + 𝑦 2 You can use complex numbers to represent a locus of points on an Argand diagram Use an algebraic method to find the Cartesian equation of the locus of z if: Represent the locus of z on an Argand diagram 𝑧−3 = 𝑧+𝑖 Replace z with x + iy 𝑥+𝑖𝑦−3 = 𝑥+𝑖𝑦+𝑖 Group real and imaginary parts 𝑥−3 +𝑖𝑦 = 𝑥+𝑖(𝑦+1) 𝑧−3 = 𝑧+𝑖 Use the rule above to remove the moduli 𝑥− 𝑦 2 = 𝑥 2 + 𝑦+1 2 𝒚=−𝟑𝒙+𝟒 Square both sides 𝑥− 𝑦 2 = 𝑥 2 + 𝑦+1 2 Expand brackets 𝑥 2 −6𝑥+9+ 𝑦 2 = 𝑥 2 + 𝑦 2 +2𝑦+1 Cancel terms −6𝑥+9=2𝑦+1 Subtract 1 −6𝑥+8=2𝑦 Divide by 2 −3𝑥+4=𝑦 3F

Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram 𝑧− 𝑧 1 Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Use an algebraic method to find the Cartesian equation of the locus of z if: Represent the locus of z on an Argand diagram y (0,4) 𝑧−3 = 𝑧+𝑖 𝒚=−𝟑𝒙+𝟒 x (3,0) (0,-1) 𝑧−3 = 𝑧+𝑖 y = -3x + 4 Distance from (3,0) Distance from (0,-1) 3F

Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram 𝑧− 𝑧 1 Further complex numbers 𝐼𝑓: 𝑧=𝑥+𝑖𝑦 𝑇ℎ𝑒𝑛: 𝑧 = 𝑥+𝑖𝑦 = 𝑥 2 + 𝑦 2 You can use complex numbers to represent a locus of points on an Argand diagram If: Use algebra to show that the locus of z is a circle, stating its centre and radius b) Sketch the locus of z on an Argand diagram 𝑧−6 =2 𝑧+6−9𝑖 Replace z with ‘x + iy’ 𝑥+𝑖𝑦−6 =2 𝑥+𝑖𝑦+6−9𝑖 Group real and imaginary parts 𝑥−6 +𝑖𝑦 =2 𝑥+6 +𝑖(𝑦−9) 𝑧−6 =2 𝑧+6−9𝑖 Replace the moduli using the rule above 𝑥− 𝑦 2 =2 𝑥 𝑦−9 2 Square both sides (remember the ‘2’) 𝑥− 𝑦 2 = 4 𝑥 𝑦−9 2 Expand some brackets 𝑥 2 −12𝑥+36+ 𝑦 2 = 𝒙+𝟏𝟎 𝟐 + 𝒚−𝟏𝟐 𝟐 =𝟏𝟎𝟎 4 𝑥 2 +12𝑥+36+ 𝑦 2 −18𝑦+81 Expand another bracket Circle, centre (-10,12) and radius 10 𝑥 2 −12𝑥+36+ 𝑦 2 = 4 𝑥 2 +48𝑥 𝑦 2 −72𝑦+324 Group all terms on one side 0= 3 𝑥 2 +60𝑥+ 3𝑦 2 −72𝑦+432 Divide by 3 0= 𝑥 2 +20𝑥+ 𝑦 2 −24𝑦+144 Completing the square 0= 𝑥 − + 𝑦−12 2 − + 144 Simplify 0= 𝑥 𝑦−12 2 −100 Add 100 100= 𝑥 𝑦−12 2 3F

Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram 𝑧− 𝑧 1 Further complex numbers y The circle shows the set of points that are twice as far from (6,0) as they are from (-6,9)! You can use complex numbers to represent a locus of points on an Argand diagram If: Use algebra to show that the locus of z is a circle, stating its centre and radius b) Sketch the locus of z on an Argand diagram (-10,12) (-6,9) 𝑧−6 =2 𝑧+6−9𝑖 P(x,y) x (6,0) 𝒙+𝟏𝟎 𝟐 + 𝒚−𝟏𝟐 𝟐 =𝟏𝟎𝟎 Circle, centre (-10,12) and radius 10 𝑧−6 =2 𝑧+6−9𝑖 The distance from (6,0) Is equal to Twice the distance from (-6,9) 3F

Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram 𝑧− 𝑧 1 Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram If: Use algebra to show that the locus of z is a circle, stating its centre and radius b) Sketch the locus of z on an Argand diagram Therefore: When 𝑧− 𝑧 1 =𝑎 𝑧− 𝑧 2  An Algebraic method will most likely be the best way to find the equation of the locus of z 𝑧−6 =2 𝑧+6−9𝑖  You will probably have to use completing the square (sometimes with fractions as well!) 𝒙+𝟏𝟎 𝟐 + 𝒚−𝟏𝟐 𝟐 =𝟏𝟎𝟎 Circle, centre (-10,12) and radius 10 3F

y You can use complex numbers to represent a locus of points on an Argand diagram If: Sketch the locus of P(x,y) which is represented by z on an Argand diagram. Then find the Cartesian equation of this locus algebraically.  The locus will be the set of points which start at (0,0) and make an argument of π/4 with the positive x-axis The line is not extended back downwards  It is known as a ‘half-line’ 𝑎𝑟𝑔𝑧= 𝜋 4 𝝅 𝟒 x 3F

y You can use complex numbers to represent a locus of points on an Argand diagram If: Sketch the locus of P(x,y) which is represented by z on an Argand diagram. Then find the Cartesian equation of this locus algebraically.  The locus will be the set of points which start at (0,0) and make an argument of π/4 with the positive x-axis y 𝝅 𝟒 x x 𝑎𝑟𝑔𝑧= 𝜋 4 𝑎𝑟𝑔𝑧= 𝜋 4 Replace z with ‘x + iy’ 𝑎𝑟𝑔(𝑥+𝑖𝑦)= 𝜋 4 The value of the argument is tan-1(opposite/adjacent) 𝑇𝑎 𝑛 −1 𝑦 𝑥 = 𝜋 4 ‘Normal tan’ 𝑦 𝑥 =𝑇𝑎𝑛 𝜋 4 Calculate the tan part 𝑦 𝑥 =1 Multiply by x 𝑦=𝑥 (x > 0) 3F

y You can use complex numbers to represent a locus of points on an Argand diagram If: Sketch the locus of P(x,y) which is represented by z on an Argand diagram. Then find the Cartesian equation of this locus algebraically. The locus will be the set of values that, when we subtract 2 from them, make an angle of π/3 with the origin The locus must therefore start at (2,0) rather than (0,0)! 𝝅 𝟑 (2,0) x arg⁡(𝑧−2)= 𝜋 3 3F

y You can use complex numbers to represent a locus of points on an Argand diagram If: Sketch the locus of P(x,y) which is represented by z on an Argand diagram. Then find the Cartesian equation of this locus algebraically. The locus will be the set of values that, when we subtract 2 from them, make an angle of π/3 with the origin The locus must therefore start at (2,0) rather than (0,0)! y 𝝅 𝟑 (2,0) x x - 2 arg⁡(𝑧−2)= 𝜋 3 arg⁡(𝑧−2)= 𝜋 3 Replace z with ‘x + iy’ 𝑎𝑟𝑔(𝑥+𝑖𝑦−2)= 𝜋 3 The value of the argument is tan-1(opposite/adjacent) 𝑇𝑎 𝑛 −1 𝑦 𝑥−2 = 𝜋 3 ‘Normal tan’ 𝑦 𝑥−2 =𝑇𝑎𝑛 𝜋 3 Calculate the tan part 𝑦 𝑥−2 = 3 Multiply by (x – 2) 𝑦= 3 𝑥−2 3 (x > 2) 3F

y You can use complex numbers to represent a locus of points on an Argand diagram If: Sketch the locus of z on an Argand diagram and use an algebraic method to find the equation of the line. When we add 3 and 2i to z, the argument from (0,0) and the positive x-axis will be 3π/4 So the line will have to start at (-3,-2) x 𝟑𝝅 𝟒 𝑎𝑟𝑔⁡(𝑧+3+2𝑖)= 3𝜋 4 (-3,-2) 3F

y You can use complex numbers to represent a locus of points on an Argand diagram If: Sketch the locus of z on an Argand diagram and use an algebraic method to find the equation of the line. When we add 3 and 2i to z, the argument from (0,0) and the positive x-axis will be 3π/4 So the line will have to start at (-3,-2) y + 2 x 𝟑𝝅 𝟒 𝑎𝑟𝑔⁡(𝑧+3+2𝑖)= 3𝜋 4 x + 3 (-3,-2) 𝑎𝑟𝑔⁡(𝑧+3+2𝑖)= 3𝜋 4 Replace z with ‘x + iy’ 𝑎𝑟𝑔(𝑥+𝑖𝑦+3+2𝑖)= 3𝜋 4 The value of the argument is tan-1(opposite/adjacent) 𝑇𝑎 𝑛 −1 𝑦+2 𝑥+3 = 3𝜋 4 ‘Normal tan’ 𝑦+2 𝑥+3 =𝑇𝑎𝑛 3𝜋 4 Calculate the tan part 𝑦+2 𝑥+3 =−1 Multiply by (x + 3) 𝑦+2=−𝑥−3 Subtract 2 (x < -3) 𝑦=−𝑥−5 3F

You can use complex numbers to represent a locus of points on an Argand diagram So therefore: Is represented by a half line starting at z1 and making an angle of θ with a line parallel to the x-axis 𝑎𝑟𝑔 𝑧− 𝑧 1 =𝜃 3F

Joining the ends of a chord to different points on the circumference will always create the same angle, if the points are in the same sector “Angles in the same sector are equal” You can use complex numbers to represent a locus of points on an Argand diagram For the next set of Loci, you need to remember some rules relating to circles Major arc – θ is acute Minor arc – θ is obtuse Semi-circle – θ is 90° θ θ x θ A B θ θ θ A B 2x A B A B If they are joined to a point on the major arc  The angle will be acute If they are joined to a point on the minor arc  The angle will be obtuse If the chord is the diameter of the circle  The angle will be 90°  The angle at the centre is twice the angle at the circumference 3F

You can use complex numbers to represent a locus of points on an Argand diagram If: Sketch the locus of P(x,y), which is represented by z on an Argand diagram  The argument above can be rewritten using this rule: 𝑎𝑟𝑔 𝑧−6 𝑧−2 = 𝜋 4 𝑎𝑟𝑔 𝑧 1 𝑧 2 =𝑎𝑟𝑔 𝑧 1 −𝑎𝑟𝑔 𝑧 2 𝑎𝑟𝑔 𝑧−6 𝑧−2 = 𝜋 4 𝑎𝑟𝑔 𝑧−6 −𝑎𝑟𝑔 𝑧−2 = 𝜋 4 So what we are doing is drawing the locus of points where the difference between these arguments is π/4 3F

y arg(z – 6) = θ1 You can use complex numbers to represent a locus of points on an Argand diagram If: Sketch the locus of P(x,y), which is represented by z on an Argand diagram arg(z – 2) = θ2 θ2 This angle must therefore be θ1 – θ2, the difference between the arguments! 𝑎𝑟𝑔 𝑧−6 𝑧−2 = 𝜋 4 θ1 θ1 θ2 𝑎𝑟𝑔 𝑧−6 −𝑎𝑟𝑔 𝑧−2 = 𝜋 4 x (2,0) (6,0) So what we are doing is drawing the locus of points where the difference between these arguments is π/4 Imagine drawing both arguments – we will use θ1 and θ2 to represent their values Using alternate angles, we can show the angle between the arguments is their difference We want this difference to be π/4 However, there are more points that satisfy this rule! 3F

y You can use complex numbers to represent a locus of points on an Argand diagram If: Sketch the locus of P(x,y), which is represented by z on an Argand diagram 𝑎𝑟𝑔 𝑧−6 𝑧−2 = 𝜋 4 π/4 π/4 θ1 θ2 θ1 θ2 𝑎𝑟𝑔 𝑧−6 −𝑎𝑟𝑔 𝑧−2 = 𝜋 4 x (2,0) (6,0) So what we are doing is drawing the locus of points where the difference between these arguments is π/4 If we move the point where the lines cross along the major arc of a circle, then the value of π/4 will remain the same The arguments will change but this doesn’t matter, it is the difference that matters! So the locus of a difference between arguments is always given by an arc of a circle Geogebra Example 3F

y “The angle at the centre is twice the angle at the circumference” You can use complex numbers to represent a locus of points on an Argand diagram If: Sketch the locus of P(x,y), which is represented by z on an Argand diagram Find the Cartesian equation of this locus We need the centre of the ‘circle’ and its radius  We need to use another of the rules we saw: π/4 𝑎𝑟𝑔 𝑧−6 𝑧−2 = 𝜋 4 π/2 x (2,0) (6,0) We can use this isosceles triangle to find the information we need… Centre Radius Radius 3F

You can use complex numbers to represent a locus of points on an Argand diagram If: Sketch the locus of P(x,y), which is represented by z on an Argand diagram Find the Cartesian equation of this locus We need the centre of the ‘circle’ and its radius Centre Radius Radius (4,2) 𝑎𝑟𝑔 𝑧−6 𝑧−2 = 𝜋 4 π/4 2 π/4 (2,0) 2 (4,0) (6,0) Split the triangle in the middle, the smaller angles will both be π/4 (45ᵒ) (because the top angle was π/2) The middle of the base will be (4,0), and you can work out the side lengths from this The top will therefore be at (4,2) Use Pythagoras’ Theorem to find the diagonal (the radius) Centre (4,2) Radius 2√2 3F

y You can use complex numbers to represent a locus of points on an Argand diagram If: Sketch the locus of P(x,y), which is represented by z on an Argand diagram Find the Cartesian equation of this locus We need the centre of the ‘circle’ and its radius π/4 𝑎𝑟𝑔 𝑧−6 𝑧−2 = 𝜋 4 x (2,0) (6,0) The locus is therefore the arc of a circle with the following equation: 𝑥− 𝑦−2 2 =8 Centre (4,2) Radius 2√2 𝑦>0 3F

You can use complex numbers to represent a locus of points on an Argand diagram Generally, for this type of question, you need to follow 3 steps: Step 1: Mark on the Argand diagram the two points where the arguments start Step 2: Decide whether the arc is going to be major, minor, or a semi-circle, by considering the angle Step 3: Draw the arc between the points. You always draw from the numerator point to the denominator point Anti-clockwise if θ is positive Clockwise if θ is negative If the value we want is positive, then θ1 > θ2 If the value we want is negative, then θ2 > θ1 Drawing in the direction indicated in step 3 means you will ensure the arguments are correct to give a positive or negative answer  As we do some examples we will refer to this! 3F

Sketch the locus of P(x,y) on an Argand diagram if: You can use complex numbers to represent a locus of points on an Argand diagram Generally, for this type of question, you need to follow 3 steps: Step 1: Mark on the Argand diagram the two points where the arguments start Step 2: Decide whether the arc is going to be major, minor, or a semi-circle, by considering the angle Step 3: Draw the arc between the points. You always draw from the numerator point to the denominator point Anti-clockwise if θ is positive Clockwise if θ is negative 𝑎𝑟𝑔 𝑧 𝑧−4𝑖 = 𝜋 2 y (0,4) (0,0) and (0,4) The angle to make is π/2  A semi-circle (0,0) x Θ is positive, so draw anti-clockwise from (0,0) (numerator point) to (0,4) (denominator point) 3F

You can use complex numbers to represent a locus of points on an Argand diagram In step 3 we had to choose whether to draw the diagram clockwise or anti-clockwise from the numerator point to the denominator point Lets show why this is correct! y y (0,4) (0,4) θ2 θ2 θ2 θ2 θ1 θ1 θ1 θ1 (0,0) x (0,0) x  We drew the angle anti-clockwise from (0,0) to (0,4)  Using the alternate angles, the angle between the arguments is θ1 + θ2  However, as θ2 is actually negative, the sum is really θ1 + (-θ2) If we drew the arc the other way - clockwise from (0.0) to (0,4) Using the alternate angles, the on the outside is θ1 + θ2  However, as θ2 is actually negative, the sum is really θ1 + (-θ2) = θ1 – θ2 But of course it is on the wrong side of the arc so we do not want this part of the circle! = θ1 – θ2  This angle is therefore what we were wanting! Basically, always use the rule in step 3! 3F

Sketch the locus of P(x,y) on an Argand diagram if: You can use complex numbers to represent a locus of points on an Argand diagram Generally, for this type of question, you need to follow 3 steps: Step 1: Mark on the Argand diagram the two points where the arguments start Step 2: Decide whether the arc is going to be major, minor, or a semi-circle, by considering the angle Step 3: Draw the arc between the points. You always draw from the numerator point to the denominator point Anti-clockwise if θ is positive Clockwise if θ is negative 𝑎𝑟𝑔 𝑧+3𝑖 𝑧−2 = 𝜋 3 y (0,-3) and (0,2) The angle to make is π/3  A major arc (0,2) x Θ is positive, so draw anti-clockwise from (0,-3) (numerator point) to (0,2) (denominator point) (0,-3) 3F

y You can use complex numbers to represent a locus of points on an Argand diagram Given that the complex number z = x + iy satisfies the equation: Find the minimum and maximum values of |z| Start by drawing this on an Argand diagram It is a circle, centre (12,5) radius 3 units 3 3 (12,5) 𝑧−12−5𝑖 =3 13 x The smallest and largest values for |z| will be on the same straight line through the circle’s centre You can mark the size of the radius on the diagram Find the distance from (0,0) to (12,5), then add/subtract 3 to find the largest and smallest values So the largest value of |z| will be 16 and the smallest will be 10 =13 3F

Teachings for exercise 3G

y You can use complex numbers to represent regions on a Argand diagram This is very similar to what you have been doing with loci The only extra part is that once you have drawn the locus representing the point, you need to indicate the area required Shade on an Argand diagram the region indicated by:  Start with a circle, centre (4,2) and radius 2 units (as 2 is the ‘limit’) (4,2) x 𝑧−4−2𝑖 ≤2 The region we want is where the absolute value of z is less than 2  This will be the region inside the circle 3G

y You can use complex numbers to represent regions on a Argand diagram This is very similar to what you have been doing with loci The only extra part is that once you have drawn the locus representing the point, you need to indicate the area required Shade on an Argand diagram the region indicated by:  Start with the perpendicular bisector between (4,0) and (6,0) as this is the ‘limit’ (4,0) (6,0) x 𝑧−4 < 𝑧−6 𝑧−4 < 𝑧−6 The distance to |z – 4| must be less than the distance to |z – 6|  Shade the region closest to (4,0) 3G

y You can use complex numbers to represent regions on a Argand diagram This is very similar to what you have been doing with loci The only extra part is that once you have drawn the locus representing the point, you need to indicate the area required Shade on an Argand diagram the region indicated by:  Start by drawing the limits of the argument from the point (2,2) 𝝅 𝟒 (2,2) x 0≤𝑎𝑟𝑔 𝑧−2−2𝑖 ≤ 𝜋 4 The argument must be between these two values  Shade the region between the two arguments 3G

y You can use complex numbers to represent regions on a Argand diagram This is very similar to what you have been doing with loci The only extra part is that once you have drawn the locus representing the point, you need to indicate the area required Shade on an Argand diagram the region indicated by: and x 𝑧−4−2𝑖 ≤2 Imagine all the regions were on the same diagram  The region we want will have to satisfy all of these at the same time! 𝑧−4 < 𝑧−6 0≤𝑎𝑟𝑔 𝑧−2−2𝑖 ≤ 𝜋 4 3G

Test Your Understanding
P6 June 2003 Q4(i)(b) Shade the region for which 𝑧−1 ≤1 and 𝜋 12 ≤ arg 𝑧+1 ≤ 𝜋 2 ?

Teachings for exercise 3H

y The z-plane You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv Effectively we take a set of points in the complex plane, transform them all and map them on a new complex plane  You will need to use Algebraic methods a lot for this as visualising the transformations can be very difficult! (uses x and y) x Transformation from one plane to the next! v The w-plane (uses u and v) u 3H

y v You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The point P represents the complex number z on an Argand diagram where |z| = 2. T1 represents a transformation from the z plane, where z = x + iy, to the w-plane where w = u + iv. Describe the locus of P under the transformation T1, when T1 is given by:  We will work out the new set of points algebraically… x u The z-plane The w-plane Circle centre (0,0), radius 2 Circle centre (-2,4), radius 2  To start with, make z the subject 𝑤=𝑧−2+4𝑖 Add 2, subtract 4i 𝑧=𝑤+2−4𝑖 The modulus of each side must be the same 𝑇 1 : 𝑤=𝑧−2+4𝑖 𝑧 = 𝑤+2−4𝑖 We know |z| from the question 2= 𝑤+2−4𝑖 Circle, centre (-2,4), radius 2 3H

The z-plane y The w-plane v You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The point P represents the complex number z on an Argand diagram where |z| = 2. T2 represents a transformation from the z plane, where z = x + iy, to the w-plane where w = u + iv. Describe the locus of P under the transformation T2, when T2 is given by:  We will work out the new set of points algebraically… x u Circle centre (0,0), radius 2 Circle centre (0,0), radius 6  To start with, make z the subject 𝑤=3𝑧 𝑤 3 =𝑧 Divide by 3 Modulus of both sides 𝑤 3 = 𝑧 𝑇 2 : 𝑤=3𝑧 |z|= 2 Split the modulus up 𝑧 1 𝑧 2 = 𝑧 𝑧 2 𝑤 3 =2 |3|=3 so multiply by 3 𝑤 =6 3H

The z-plane y The w-plane v You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The point P represents the complex number z on an Argand diagram where |z| = 2. T2 represents a transformation from the z plane, where z = x + iy, to the w-plane where w = u + iv. Describe the locus of P under the transformation T3, when T3 is given by:  We will work out the new set of points algebraically… x u Circle centre (0,0), radius 2 Circle centre (0,1), radius 1  To start with, make z the subject 𝑤= 1 2 𝑧+𝑖 Leaving z like this can make the problem easier! (rather than rearranging completely) Subtract i 𝑤−𝑖= 1 2 𝑧 Modulus of both sides 𝑤−𝑖 = 1 2 𝑧 𝑇 3 : 𝑤= 1 2 𝑧+𝑖 You can split the modulus on the right 𝑧 1 𝑧 2 = 𝑧 1 𝑧 2 𝑤−𝑖 = 𝑧 |z| = 2 𝑤−𝑖 = Simplify the right side 𝑤−𝑖 =1 3H

The z-plane y The w-plane v You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv For the transformation w = z2, where z = x + iy and w = u + iv, find the locus of w when z lies on a circle with equation x2 + y2 = 16 It is very important for this topic that you draw information on z or |z| from the question The equation x2 + y2 = 16 is a circle, centre (0,0) and radius 4 Therefore |z| = 4 We now proceed as before, by writing the equation linking w and z in such a way that |z| can be replaced x u Circle centre (0,0), radius 4 Circle centre (0,0), radius 16 𝑤= 𝑧 2 Modulus of both sides 𝑤 = 𝑧 2 Split the modulus up 𝑤 = 𝑧 𝑧 Replace |z| with 4 𝑤 =4×4 Calculate 𝑤 =16 Circle, centre (0,0) and radius 16 3H

𝑤= 5𝑖𝑧+𝑖 𝑧+1 You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: Show that the image, under T, of the circle |z| = 1 in the z-plane, is a line l in the w-plane. Sketch l on an Argand diagram. Make z the subject! Now eliminate z using what we know… Multiply by (z + 1) 𝑤(𝑧+1)=5𝑖𝑧+𝑖 Expand the bracket 𝑤𝑧+𝑤=5𝑖𝑧+𝑖 Subtract 5iz and subtract w 𝑤𝑧−5𝑖𝑧=𝑖−𝑤 Factorise the left side 𝑤= 5𝑖𝑧+𝑖 𝑧+1 𝑧(𝑤−5𝑖)=𝑖−𝑤 Divide by (w – 5i) 𝑧= 𝑖−𝑤 𝑤−5𝑖 Modulus of both sides 𝑧 = 𝑖−𝑤 𝑤−5𝑖 |z| = 1 1= 𝑖−𝑤 𝑤−5𝑖 Multiply by |w – 5i| 𝑤−5𝑖 = 𝑖−𝑤 |i - w| = |w – i| 𝑤−5𝑖 = 𝑤−𝑖 3H

Transformation T 𝑧 =1 𝑤−5𝑖 = 𝑤−𝑖 You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: Show that the image, under T, of the circle |z| = 1 in the z-plane, is a line l in the w-plane. Sketch l on an Argand diagram. Circle centre (0,0), radius 1 Perpendicular bisector between (0,1) and (0,5)  The line v = 3 The z-plane y The w-plane v 𝑤= 5𝑖𝑧+𝑖 𝑧+1 x u So a circle can be transformed into a straight line! 3H

Test Your Understanding
FP2 June 2009 Q6 A transformation 𝑇 from the 𝑧-plane to the 𝑤-plane is given by 𝑤= 𝑧 𝑧+𝑖 , 𝑧≠−𝑖 The circle with equation 𝑧 =3 is mapped by 𝑇 onto the curve 𝐶. Show that 𝐶 is a circle and find its centre and radius. (8) The region 𝑧 <3 in the 𝑧-plane is mapped by 𝑇 onto the region 𝑅 in the 𝑤-plane. Shaded the region 𝑅 on an Argand diagram. (2) (Hint: be careful and don’t guess! Try a value of 𝑧 where you know that 𝑧 <3, and then transform it to see where it ends up.) ? ?

𝑤= 3𝑧−2 𝑧+1 You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: Show that the image, under T, of the circle with equation x2 + y2 = 4 in the z-plane, is a different circle C in the w-plane. State the centre and radius of C.  Remember that x2 + y2 = 4 is the same as |z| = 2 Multiply by (z + 1) 𝑤 𝑧+1 =3𝑧−2 Expand the bracket 𝑤𝑧+𝑤=3𝑧−2 Subtract wz and add 2 𝑤+2=3𝑧−𝑤𝑧 Factorise the right side 𝑤+2=𝑧(3−𝑤) 𝑤= 3𝑧−2 𝑧+1 Divide by (3 – w) 𝑤+2 3−𝑤 =𝑧 Modulus of each side 𝑤+2 3−𝑤 = 𝑧 Split up the modulus |z| = 2 𝑤+2 3−𝑤 =2 Multiply by |3 - w| 𝑤+2 =2 3−𝑤 |3 - w| = |w - 3| 𝑤+2 =2 𝑤−3 We now need to find what the equation of this will be! 3H

We will find the equation as we did in the early part of section 3F! Further complex numbers 𝑤+2 =2 𝑤−3 You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: Show that the image, under T, of the circle with equation x2 + y2 = 4 in the z-plane, is a different circle C in the w-plane. State the centre and radius of C.  Remember that x2 + y2 = 4 is the same as |z| = 2 Replace w with ‘u + iv’ 𝑢+𝑖𝑣+2 =2 𝑢+𝑖𝑣−3 Group real/imaginary terms 𝑢+2 +𝑖𝑣 =2 (𝑢−3)+𝑖𝑣 Remove the modulus 𝑢 𝑣 2 = 4 𝑢− 𝑣 2 Expand brackets 𝑢 2 +4𝑢+4+ 𝑣 2 = 4 𝑢 2 −6𝑢+9+ 𝑣 2 Expand more brackets! 𝑤= 3𝑧−2 𝑧+1 𝑢 2 +4𝑢+4+ 𝑣 2 = 4𝑢 2 −24𝑢+36+ 4𝑣 2 Move all to one side 0= 3𝑢 2 −28𝑢+3 𝑣 2 +32 0= 𝑢 2 − 28 3 𝑢+ 𝑣 Divide by 3 Use completing the square 0= 𝑢− − + 𝑣 Move the number terms across 100 9 = 𝑢− 𝑣 2 Circle, centre (14/3,0), radius 10/3 3H

𝑧 =2 Transformation T 𝑤+2 =2 𝑤−3 You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: Show that the image, under T, of the circle with equation x2 + y2 = 4 in the z-plane, is a different circle C in the w-plane. State the centre and radius of C. 𝑢− 𝑣 2 = 100 9 𝑥 2 + 𝑦 2 =4 Circle centre (0,0), radius 2 Circle, centre (14/3,0), radius 10/3 y v 𝑤= 3𝑧−2 𝑧+1 The z-plane The w-plane x u 3H

𝑤= 𝑖𝑧−2 1−𝑧 You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv A transformation T of the z-plane to the w-plane is given by: Show that as z lies on the real axis in the z-plane, then w lies on a line l in the w-plane. Find the equation of l and sketch it on an Argand diagram.  Start by rearranging to make z the subject ‘as usual’ Multiply by (1 – z) 𝑤 1−𝑧 =𝑖𝑧−2 Expand the bracket 𝑤−𝑤𝑧=𝑖𝑧−2 Add 2, Add wz 𝑤+2=𝑖𝑧+𝑤𝑧 Factorise the right side 𝑤+2=𝑧(𝑖+𝑤) 𝑤= 𝑖𝑧−2 1−𝑧 Divide by (i + w) 𝑤+2 𝑖+𝑤 =𝑧 Write the other way round (if you feel it is easier!) 𝑧= 𝑤+2 𝑖+𝑤 i + w = w + i 𝑧= 𝑤+2 𝑤+𝑖 At this point we have a problem, as we do not know anything about |z| However, as z lies on the ‘real’ axis, we know that y = 0 Replace z with ‘x + iy’ 𝑥+𝑖𝑦= 𝑤+2 𝑤+𝑖 3H

𝑥+𝑖𝑦= 𝑤+2 𝑤+𝑖 You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv A transformation T of the z-plane to the w-plane is given by: Show that as z lies on the real axis in the z-plane, then w lies on a line l in the w-plane. Find the equation of l and sketch it on an Argand diagram. Now, you need to rewrite the right side so you can separate all the real and imaginary terms You must be extremely careful with positives and negatives here! Replace w with u + iv 𝑥+𝑖𝑦= 𝑢+𝑖𝑣+2 𝑢+𝑖𝑣+𝑖 Group real and imaginary terms 𝑥+𝑖𝑦= 𝑢+2 +𝑖𝑣 𝑢+𝑖(𝑣+1) Multiply by the denominator but with the opposite sign (this will cancel ‘i’ terms on the bottom 𝑤= 𝑖𝑧−2 1−𝑧 𝑥+𝑖𝑦= 𝑢+2 +𝑖𝑣 𝑢+𝑖(𝑣+1) × 𝑢−𝑖(𝑣+1) 𝑢−𝑖(𝑣+1) 𝑢(𝑢+2) − 𝑖(𝑢+2)(𝑣+1) + 𝑖𝑢𝑣 − 𝑖 2 𝑣(𝑣+1) 𝑥+𝑖𝑦= 𝑢 2 − 𝑖𝑢(𝑣+1) + 𝑖𝑢(𝑣+1) − 𝑖 2 (𝑣+1) 2 Simplify i2 = -1 𝑢(𝑢+2) + 𝑣(𝑣+1) + 𝑖𝑢𝑣 − 𝑖(𝑢+2)(𝑣+1) 𝑥+𝑖𝑦= 𝑢 2 + (𝑣+1) 2 Separate real and ‘i’ terms 𝑢 𝑢+2 +𝑣(𝑣+1) 𝑢 2 + 𝑣 𝑖 𝑢𝑣−(𝑢+2)(𝑣+1) 𝑢 2 + 𝑣+1 2 𝑥+𝑖𝑦= As z lies on the x-axis, we know y = 0  Therefore, the imaginary part on the right side must also equal 0 3H

𝑢 𝑢+2 +𝑣(𝑣+1) 𝑢 2 + 𝑣 𝑖 𝑢𝑣−(𝑢+2)(𝑣+1) 𝑢 2 + 𝑣+1 2 𝑥+𝑖𝑦= You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv A transformation T of the z-plane to the w-plane is given by: Show that as z lies on the real axis in the z-plane, then w lies on a line l in the w-plane. Find the equation of l and sketch it on an Argand diagram. You can now find an equation for the line in the w-plane  Set the imaginary part equal to 0 𝑢𝑣−(𝑢+2)(𝑣+1) 𝑢 2 + 𝑣 =0 Multiply by (u2 + (v + 1)2) (you will be left with the numerator) 𝑢𝑣−(𝑢+2)(𝑣+1)=0 𝑤= 𝑖𝑧−2 1−𝑧 Multiply out the double bracket 𝑢𝑣−(𝑢𝑣+2𝑣+𝑢+2)=0 Subtract all these terms 𝑢𝑣−𝑢𝑣−2𝑣−𝑢−2=0 The ‘uv’ terms cancel out −2𝑣−𝑢−2=0 Make v the subject 2𝑣=−𝑢−2 Divide by 2 𝑣=− 1 2 𝑢−1 So the transformation has created this line in the w-plane (remember v is essentially ‘y’ and u is essentially ‘x’) 3H

Transformation T You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv A transformation T of the z-plane to the w-plane is given by: Show that as z lies on the real axis in the z-plane, then w lies on a line l in the w-plane. Find the equation of l and sketch it on an Argand diagram. 𝑣=− 1 2 𝑢−1 𝑦=0 (z lies on the real axis) Straight line, gradient is 1/2 and y-intercept at (0,-1) 𝑤= 𝑖𝑧−2 1−𝑧 y v The z-plane The w-plane x u 𝒚=𝟎 -2 -1 𝒗=− 𝟏 𝟐 𝒖−𝟏 3H

Summary You have learnt a lot in this chapter!!
You have seen proofs of and uses of De Moivre’s theorem You have found real and complex roots of powers You have see how to plot Loci and perform transformations of complex functions

Further complex numbers

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Further complex numbers

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