1. FP2
Complex numbers 3c

2. FP2: Complex numbers 3c Starter: KUS objectives
BAT use De Moivres theorem to find the nth roots of a complex number
Starter:

3. We need to apply the following results
Notes 1
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

4. k = 2 would cause the argument to be outside the range
WB 18 part1 Solve the equation z3 = 1 and represent your solutions on an Argand diagram.
𝐼𝑓: 𝑧=𝑟 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃
𝑇ℎ𝑒𝑛: 𝑧=𝑟 cos⁡(𝜃+2𝑘𝜋)+𝑖𝑠𝑖𝑛(𝜃+2𝑘𝜋)
𝑟(𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃) 𝑛 = 𝑟 𝑛 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃
First you need to express z in the modulus-argument form. Use an Argand diagram. 1 x y
𝑟=1
𝜃=0
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
𝑧=− 1 2 −𝑖
TA DA! These are the roots of z3 = 1
k = 2 would cause the argument to be outside the range

5. 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
− 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

6. 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 − 𝑟= 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
𝑧= 𝑐𝑜𝑠 𝜋 3 +2𝑘𝜋 4 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝑘𝜋 4
𝑧= 2 𝑐𝑜𝑠 𝜋 3 +2𝑘𝜋 4 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝑘𝜋 4

7. 𝑧= 2 𝑒 𝜋 12 𝑖 𝑧= 2 𝑒 − 5𝜋 12 𝑖 𝑧= 2 𝑒 7𝜋 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 𝑖𝑠𝑖𝑛 𝜋 3 4
Solutions in the modulus-argument form
k = 0
𝑧= 2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12
𝑧= 2 𝑐𝑜𝑠 𝜋 12 +𝑖𝑠𝑖𝑛 𝜋 12
𝑧= 2 𝑐𝑜𝑠 − 5𝜋 12 +𝑖𝑠𝑖𝑛 − 5𝜋 12
𝑧= 2 𝑐𝑜𝑠 𝜋 3 +2𝜋 4 +𝑖𝑠𝑖𝑛 𝜋 3 +2𝜋 4
𝑧= 2 𝑐𝑜𝑠 7𝜋 12 +𝑖𝑠𝑖𝑛 7𝜋 12
𝑧= 2 𝑐𝑜𝑠 − 11𝜋 12 +𝑖𝑠𝑖𝑛 − 11𝜋 12
k = 1
𝑧= 2 𝑐𝑜𝑠 7𝜋 12 +𝑖𝑠𝑖𝑛 7𝜋 12
Solutions in the exponential form
𝑧= 2 𝑒 𝜋 12 𝑖
𝑧= 2 𝑒 − 5𝜋 12 𝑖
𝑧= 2 𝑐𝑜𝑠 𝜋 3 −2𝜋 4 +𝑖𝑠𝑖𝑛 𝜋 3 −2𝜋 4
k = -1
𝑧= 2 𝑒 7𝜋 12 𝑖
𝑧= 2 𝑒 − 11𝜋 12 𝑖
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

8. 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 = cos 5𝜋 3 +𝑖 sin 5𝜋 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

9. 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 –

10. END