Complex numbers nth roots

FP2: Complex numbers : nth roots KUS objectives BAT use De Moivres theorem to find the nth roots of a complex number Starter: write in exponential form 𝑧 1 =7 2 𝑒 𝜋𝑖 4 𝑧 1 =7−7𝑖 𝑧 2 =2 cos 3𝜋 4 +𝑖 sin 11𝜋 4 𝑧 2 =2 𝑒 3𝜋𝑖 4 𝑧 3 = 3 cos 𝜋 3 +𝑖 sin 7𝜋 3 𝑧 3 = 3 𝑒 − 𝜋𝑖 3

Then: 𝑧=𝑟 cos⁡(𝜃+2𝑘𝜋)+𝑖𝑠𝑖𝑛(𝜃+2𝑘𝜋) where k is an integer Notes 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: 𝑧=𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 Then: 𝑧=𝑟 cos⁡(𝜃+2𝑘𝜋)+𝑖𝑠𝑖𝑛(𝜃+2𝑘𝜋) 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 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃

k = 2 would cause the argument to be outside the range WB 18 part1 a) Solve the equation z3 = 1 and b) represent your solutions on an Argand diagram c) Show that the three roots can be written as 1, 𝑤, 𝑎𝑛𝑑 𝑤 2 where 1+𝑤+ 𝑤 2 =0 𝐼𝑓: 𝑧=𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 𝑇ℎ𝑒𝑛: 𝑧=𝑟 cos⁡(𝜃+2𝑘𝜋)+𝑖𝑠𝑖𝑛(𝜃+2𝑘𝜋) 𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 1 x y 𝑟=1 𝜃=0 First you need to express z in the modulus-argument form. 𝑧 3 =1 𝑐𝑜𝑠0+𝑖𝑠𝑖𝑛0 Now then find an expression for z in terms of k 𝑧 3 = cos⁡(0+2𝑘𝜋)+𝑖𝑠𝑖𝑛(0+2𝑘𝜋) We can then solve this to find the roots of the equation above 𝑧= cos 0+2𝑘𝜋 +𝑖𝑠𝑖𝑛(0+2𝑘𝜋) 1 3 𝑧= cos 0+2𝑘𝜋 3 +𝑖𝑠𝑖𝑛 0+2𝑘𝜋 3 k = 0 k = 1 k = -1 𝑧= 𝑐𝑜𝑠 2𝜋 3 +𝑖𝑠𝑖𝑛 2𝜋 3 𝑧= 𝑐𝑜𝑠 − 2𝜋 3 +𝑖𝑠𝑖𝑛 − 2𝜋 3 𝑧= 𝑐𝑜𝑠 0 +𝑖𝑠𝑖𝑛 0 𝑧=1 𝑧=− 1 2 +𝑖 3 2 𝑧=− 1 2 −𝑖 3 2 TA DA! These are the roots of z3 = 1 k = 2 would cause the argument to be outside the range

WB 18 part2 Solve the equation z3 = 1 and represent your solutions on an Argand diagram 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 c) Show that the three roots can be written as 1, 𝑤, 𝑎𝑛𝑑 𝑤 2 where 1+𝑤+ 𝑤 2 =0 − 𝟏 𝟐 −𝒊 𝟑 𝟐 𝑤=− 1 2 +𝑖 3 2 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 SO the points make the vertices of a regular shape 𝑤 2 = − 1 2 +𝑖 3 2 − 1 2 +𝑖 3 2 𝑤 2 =− 1 2 −𝑖 3 2 QED 1+𝑤+ 𝑤 2 =1+ − 1 2 +𝑖 3 2 + − 1 2 −𝑖 3 2 =0 QED

In general solutions to 𝑧 4 =1 are Notes In general solutions to 𝑧 4 =1 are 𝑧= cos 2𝑘𝜋 𝑛 +𝑖 sin 2𝑘𝜋 𝑛 for k = 1, 2, 3, …, n These are known as the nth roots of unity In exponential form 𝑧= 𝑒 2𝑘𝜋 𝑛 𝑖   The nth roots are 1, 𝑤, 𝑤 2 , 𝑤 3 , … 𝑤 𝑛−1 and they form the vertices of a regular polygon Also 1+ 𝑤+ 𝑤 2 + 𝑤 3 + … + 𝑤 𝑛−1 =0

WB 19 part1 Solve the equation z4 - 2√3i = 2 Give your answers in both the modulus-argument and exponential forms. By rearranging… z4 = 2 + 2√3i 𝜃= 𝑡𝑎𝑛 −1 2 3 2 𝑟= 2 2 + 2 3 2 r 2 θ x y 2√3 𝑟=4 𝜃= 𝜋 3 As before, use an Argand diagram to express the equation in the modulus-argument form 𝑧 4 =4 𝑐𝑜𝑠 𝜋 3 +𝑖𝑠𝑖𝑛 𝜋 3 𝑧 4 =4 𝑐𝑜𝑠 𝜋 3 +2𝑘𝜋 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝑘𝜋 𝑧= 4 𝑐𝑜𝑠 𝜋 3 +2𝑘𝜋 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝑘𝜋 1 4 𝑧= 4 1 4 𝑐𝑜𝑠 𝜋 3 +2𝑘𝜋 4 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝑘𝜋 4 𝑧= 2 𝑐𝑜𝑠 𝜋 3 +2𝑘𝜋 4 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝑘𝜋 4

𝑧= 2 𝑒 𝜋 12 𝑖 𝑧= 2 𝑒 7𝜋 12 𝑖 𝑧= 2 𝑒 − 5𝜋 12 𝑖 𝑧= 2 𝑒 − 11𝜋 12 𝑖 WB 19 part2 Solve the equation z4 - 2√3i = 2 Give your answers in both the modulus-argument and exponential forms. 𝑧= 2 𝑐𝑜𝑠 𝜋 3 4 +𝑖𝑠𝑖𝑛 𝜋 3 4 Solutions in the modulus-argument form 𝑧= 2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12 𝑧= 2 𝑐𝑜𝑠 7𝜋 12 +𝑖𝑠𝑖𝑛 7𝜋 12 𝑧= 2 𝑐𝑜𝑠 − 5𝜋 12 +𝑖𝑠𝑖𝑛 − 5𝜋 12 𝑧= 2 𝑐𝑜𝑠 − 11𝜋 12 +𝑖𝑠𝑖𝑛 − 11𝜋 12 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 Solutions in the exponential form 𝑧= 2 𝑒 𝜋 12 𝑖 𝑧= 2 𝑒 7𝜋 12 𝑖 𝑧= 2 𝑒 − 5𝜋 12 𝑖 𝑧= 2 𝑒 − 11𝜋 12 𝑖 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 − 11𝜋 12 +𝑖𝑠𝑖𝑛 − 11𝜋 12

By rearranging… 𝑧 3 =−4 2 −4𝑖 2 𝑟 3 = 8 → 𝑟= 2 3𝑖𝜃=− 3𝜋 4 𝑖+2𝑘𝜋 WB 20 Solve the equation 𝑧 3 +4 2 +4𝑖 2 =0 r 4 2 θ x y 𝑟=8 𝜃=− 3𝜋 4 𝑟= −4 2 2 + −4 2 2 𝜃=−𝜋− 𝑡𝑎𝑛 −1 1 By rearranging… 𝑧 3 =−4 2 −4𝑖 2 As before, use an Argand diagram to express the equation in exponential form 𝑧 3 =8 𝑐𝑜𝑠 − 3𝜋 4 +𝑖𝑠𝑖𝑛 − 3𝜋 4 =8 𝑒 − 3𝜋 4 𝑖 𝑧 3 =8 𝑒 − 3𝜋 4 𝑖+2𝑘𝜋 𝑟 𝑒 𝑖𝜃 3 = 8 𝑒 − 3𝜋 4 𝑖+2𝑘𝜋 𝑟 3 𝑒 3𝑖𝜃 = 8 𝑒 − 3𝜋 4 𝑖+2𝑘𝜋 Equating the modulus and argument on both sides gives 𝜃=− 𝜋 4 gives 𝑧 1 =2 𝑒 − 𝜋 4 𝑖 k = 0 =2 𝑐𝑜𝑠 − 𝜋 4 +𝑖𝑠𝑖𝑛 − 𝜋 4 =2 𝑐𝑜𝑠 5𝜋 12 +𝑖𝑠𝑖𝑛 5𝜋 12 =2 𝑐𝑜𝑠 −11𝜋 12 +𝑖𝑠𝑖𝑛 −11𝜋 12 𝑟 3 = 8 → 𝑟= 2 3𝑖𝜃=− 3𝜋 4 𝑖+2𝑘𝜋 𝜃= 5𝜋 12 gives 𝑧 2 =2 𝑒 5𝜋 12 𝑖 k = 1 𝜃=− 11𝜋 12 gives 𝑧 3 =2 𝑒 − 11𝜋 12 𝑖 k = -1

Notes The nth roots of any complex number lie at the vertices of a regular polygon (n sides) with its centre at the origin We find the vertices by finding a single vertex and rotatin that point around the origin. This is equivalent to multiplying by the nth roots of unity to   If 𝑧 1 is one root of the equation and 1, 𝑤, 𝑤 2 , 𝑤 3 , … 𝑤 𝑛−1 are the nth roots of unity Then the roots are  𝑧 1 , 𝑧 1 𝑤, 𝑧 1 𝑤 2 , 𝑧 1 𝑤 3 , … , 𝑧 1 𝑤 𝑛−1

WB 21 The point 𝑃( 3 , 1) lies at one vertex of an equilateral triangle. The centre of the triangle is at the origin. Find the coordinates of the other vertices of the triangle Find the area of the triangle a) The cube roots of unity are 1, 𝑤, 𝑤 2 𝑤ℎ𝑒𝑟𝑒 𝑤= 𝑒 2𝜋𝑖 3 O x y 3 , 1 0, −2 − 3 , 1 3 +𝑖=2 cos 𝜋 6 +𝑖 sin 𝜋 6 =2 𝑒 𝜋𝑖 6 So the vertices are at 𝑧 1 , 𝑧 1 𝑤, 𝑧 1 𝑤 2 𝑧 1 =2 𝑒 𝜋𝑖 6 𝑧 1 𝑤= 2 𝑒 𝜋𝑖 6 𝑒 2𝜋𝑖 3 =2 𝑒 5𝜋𝑖 6 = − 3 +𝑖 𝑧 1 𝑤 2 = 2 𝑒 𝜋𝑖 6 𝑒 2𝜋𝑖 3 2 =2 𝑒 9𝜋𝑖 6 =2 𝑒 −𝜋𝑖 2 =−2𝑖 The coordinates are shown on the Argand diagram b) Area = 1 2 ×2 3 ×3=3 3

Summary points Make sure you get the statement of De Moivre’s theorem right De Moivre’s theorem says (𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = cos 𝑛𝜃+𝑖 sin 𝑛𝜃 that for all integers n. It does not say, for example, that 𝑐𝑜𝑠 𝑛 𝜃+i 𝑠𝑖𝑛 𝑛 𝜃= cos 𝑛𝜃+𝑖 sin 𝑛𝜃 for all integers n. This is just one of numerous possible silly errors. 2. Remember to deal with the modulus when using De Moivre’s theorem to find a power of a complex number For example in 3( cos 𝜋 3 +𝑖𝑠𝑖𝑛 𝜋 3 ) 5 = 3 5 cos 5𝜋 3 +𝑖 sin 5𝜋 3 a common mistake is to forget to raise 3 to the power of 5. 3. Make sure that you don’t get the modulus of an nth root of a complex number wrong Remember that 𝑧 𝑛 = 𝑧 𝑛 , and this applies not just to integer values of n, but includes rational values of n, as when taking roots of z. 4. Make sure that you get the right number of nth roots of a complex number There should be exactly n of them. Remember two complex numbers which have the same moduli and arguments which differ by a multiple of 2𝜋 are actually the same number

One thing to improve is – KUS objectives BAT use De Moivres theorem to find the nth roots of a complex number self-assess One thing learned is – One thing to improve is –

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