Complex numbers A2

Complex numbers: multiply and divide KUS objectives BAT know how multiplying and dividing affects both the modulus and argument of the resulting complex number Starter: Use the trig addition formula to expand and simplify sin 𝑥+30 𝑐𝑜𝑠 𝑥−45 𝑐𝑜𝑠 2𝑥+3𝑦

Notes To be able to do this you need to be able to use the addition formulas for sine and cosine 𝑠𝑖𝑛 𝜃 1 ± 𝜃 2 =𝑠𝑖𝑛 𝜃 1 𝑐𝑜𝑠 𝜃 2 ±𝑐𝑜𝑠 𝜃 1 𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 𝜃 1 ± 𝜃 2 =𝑐𝑜𝑠 𝜃 1 𝑐𝑜𝑠 𝜃 2 ∓𝑠𝑖𝑛 𝜃 1 𝑠𝑖𝑛 𝜃 2 𝑐𝑜𝑠 2 𝜃+ 𝑠𝑖𝑛 2 𝜃=1

Notes 2 Multiplying a complex number z1 by another complex number z2, both in the modulus-argument form 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 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 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

Multiply to cancel terms on the denominator Notes 4 Dividing a complex number z1 by another complex number z2, both in the modulus-argument form 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 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 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 Multiply to cancel terms on the denominator 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 Multiply out Remove i2 𝑧 1 𝑧 2 = 𝑟 1 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 2 + 𝑠𝑖𝑛 2 𝜃 2 Group real and complex 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 + 𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2 𝑧 1 𝑧 2 = Rewrite (again!) 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑐𝑜𝑠 𝜃 1 − 𝜃 2 +𝑖𝑠𝑖𝑛 𝜃 1 − 𝜃 2 Rewrite terms

WB 5 a) Express the following calculation in the form x + iy: 3 𝑐𝑜𝑠 5𝜋 12 +𝑖𝑠𝑖𝑛 5𝜋 12 ×4 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12 3 𝑐𝑜𝑠 5𝜋 12 +𝑖𝑠𝑖𝑛 5𝜋 12 ×4 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12 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 Calculate the cos and sin parts (in terms of i where needed) 12 0+𝑖(1) 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 ) Multiply out =12𝑖

WB 5 b) Express the following calculation in the form x + iy: 2 𝑐𝑜𝑠 𝜋 15 +𝑖𝑠𝑖𝑛 𝜋 15 ×3 𝑐𝑜𝑠 2𝜋 5 −𝑖𝑠𝑖𝑛 2𝜋 5 2 𝑐𝑜𝑠 𝜋 15 +𝑖𝑠𝑖𝑛 𝜋 15 ×3 𝑐𝑜𝑠 2𝜋 5 −𝑖𝑠𝑖𝑛 2𝜋 5 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 Simplify 6 𝑐𝑜𝑠 − 𝜋 3 +𝑖𝑠𝑖𝑛 − 𝜋 3 Calculate the cos and sin parts 6 1 2 +𝑖 − 3 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 ) 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 ) Multiply out =3−3 3 𝑖

WB 5 c) Express the following calculation in the form x + iy: 2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12 2 𝑐𝑜𝑠 5𝜋 6 +𝑖𝑠𝑖𝑛 5𝜋 6 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 2 𝑐𝑜𝑠 − 3𝜋 4 +𝑖𝑠𝑖𝑛 − 3𝜋 4 You can work out the sin and cos parts 2 2 − 1 2 +𝑖 − 1 2 Multiply out =− 1 2 − 1 2 𝑖 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 )

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 Notes 3 Multiplying a complex number z1 by another complex number z2, both in the exponential form Multiplying a complex number z1 by another complex number z2, both in the exponential form 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 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖( 𝜃 1 + 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

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 = Notes 5 Dividing a complex number z1 by another complex number z2, both in the exponential form 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 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 𝜃 1 −𝑖 𝜃 2 Factorise the power 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 𝑒 𝑖 (𝜃 1 − 𝜃 2 ) 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

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 ) WB 6 Express the following calculations in the form x + iy: 𝑎) 2 𝑒 𝜋𝑖 6 × 3 𝑒 𝜋𝑖 3 𝑏) 2 𝑒 𝜋𝑖 3 3 𝑒 𝜋𝑖 6 𝑎) 2× 3 𝑒 𝜋𝑖 6 + 𝜋𝑖 3 =2 3 𝑒 𝜋𝑖 2 =2 3 cos 𝜋 2 +𝑖𝑠𝑖𝑛 𝜋 2 =2 3 0+𝑖 = 2 3 𝑖 or 2𝑖 3 𝑏) 2 𝑒 𝜋𝑖 3 3 𝑒 𝜋𝑖 6 = 2 3 𝑒 𝜋𝑖 3 − 𝜋𝑖 6 = 2 3 3 𝑒 𝜋𝑖 6 = 2 3 3 cos 𝜋 6 +𝑖𝑠𝑖𝑛 𝜋𝑖 6 = 2 3 3 3 2 + 1 2 𝑖 = 1+ 3 3 i

2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12 2 𝑐𝑜𝑠 5𝜋 6 +𝑖𝑠𝑖𝑛 5𝜋 6 = 2 𝑒 𝜋𝑖/12 2 𝑒 5𝜋𝑖/6 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 ) WB 7 Express 2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12 2 𝑐𝑜𝑠 5𝜋 6 +𝑖𝑠𝑖𝑛 5𝜋 6 in the form r 𝑒 𝑖𝜃 2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12 2 𝑐𝑜𝑠 5𝜋 6 +𝑖𝑠𝑖𝑛 5𝜋 6 = 2 𝑒 𝜋𝑖/12 2 𝑒 5𝜋𝑖/6 = 2 2 𝑒 𝑖 𝜋 12 − 5𝜋 6 = 2 𝑒 3𝜋𝑖 4

so w=3 𝑒 −𝜋𝑖/4 or w=3 𝑒 3𝜋𝑖/4 1st 𝑧𝑤 =3 𝑧 we know 𝑧𝑤 = 𝑧 𝑤 so 𝑤 =3 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 ) WB 8 𝑧=2+2𝑖, 𝐼𝑚 𝑧𝑤 =0 𝑎𝑛𝑑 𝑧𝑤 =3 𝑧 use geometrical reasoning to find the two possibilities for w, giving them in exponential form 1st 𝑧𝑤 =3 𝑧 we know 𝑧𝑤 = 𝑧 𝑤 so 𝑤 =3 𝑧𝑤 2 𝑧𝑤 1 𝑧𝑤 lies on the real axis 𝑅𝑒 𝐼𝑚 2nd arg 𝑧 = arctan 2 2 = 𝜋 4 𝑧=2+2𝑖 we know arg zw =arg z+arg 𝑤 3𝜋 4 𝜋 4 im (𝑧𝑤) =0 so arg (𝑧𝑤) =0 or 𝜋 so arg 𝑤 =− 𝜋 4 or 3𝜋 4 so 𝑧 is rotated − 𝜋 4 cw or 3𝜋 4 acw When multiplied by w so w=3 𝑒 −𝜋𝑖/4 or w=3 𝑒 3𝜋𝑖/4

KUS objectives BAT know how multiplying and dividing affects both the modulus and argument of the resulting complex number self-assess One thing learned is – One thing to improve is –

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