Complex Numbers

Question 1

Exponential form from mod arg form for complex numbers

Answer 1

re^iθ = r(cos θ + i sin θ)

Question 2

Complex to exponential

Answer 2

Find the mod (r) and the argument (θ) and substitute into the exponential format

Question 3

Exponential to complex

Answer 3

Put into modulus-argument form and then into the complex form

Question 4

Multiply and divide in exponential form

Answer 4

Multiply/divide the mods and add/subtract the args before writing in exponential form

Question 5

When to simplify mod-arg

Answer 5

When there are simple values for θ that give exact values e.g. π/2

Question 6

De Moivre’s theorem
(r(cos(θ) + isin(θ)))^n

Answer 6

r^n(cos(nθ) + isin(nθ))

Question 7

Proving de Moivre’s theorem by induction

Answer 7

do z^k x z and simplify to r^k+1(cos(k+1)θ + isin(k+1)θ) using the addition formulae

Question 8

de Moive’s theorem
re^iθ

Answer 8

n inθ
r^e^

Question 9

(x + yi)^n

Answer 9

Write x + yi in mod-arg and apply de Moivre’s theorem

Question 10

Express cos/sin nθ in terms of powers of cos/sin nθ

Answer 10

1. Use de Moivre’s theorem with n as the power
2. Expand (cosθ + isinθ)^n using binomial expansion
3. Set the real/imaginary part of each side equal
4. Simplify to be in terms of cos/sin

Question 11

1/2(z^nθi + z^-nθi)

Answer 11

cosnθ

Question 12

1/2i(z^nθi - z^-nθi)

Answer 12

sin(nθ)

Question 13

cos/sin^nθ in terms of cos/sin nθ

Answer 13

1. Use (2cosθ)^n = (z + 1/z)^n or (2sinθ) = (z - 1/z)^n
2. Expand both sides remembering the 2/2i
3. Group the RHS with z^n +/- 1/z^n
4. Use the identities for (z +/- 1/z) and substitute
5. Divide both sides by the coefficient on the LHS

If it has both cos and sin expand both terms and multiply

Question 14

Showing z^n +/- z^-n is 2cosθ or 2isinθ

Answer 14

Use de Moivre’s theorem with n and -n, simplifying the negative signs

Question 15

n-1
∑ w z^r
r=0

Answer 15

w(z^n-1)/(z - 1)

Question 16

∞
∑ w z^r if |z| < 1
r=0

Answer 16

w/(z-1)

Question 17

Simplifying sums to an-1

Answer 17

1. Write out the sum of series rule
2. Replace w with the first term and z with its exponential form, using e^πi = -1
3. Multiply top and bottom by e^(-1/2 x power in denominator)
4. Use the 2cosθ and 2isinθ rules in the denominator
5. Multiply top and bottom by i, use I^2 = -1 and simplify
6. Put numerator in mod-arg and simplify to the needed form

Question 18

e^πi

Answer 18

-1

Question 19

z^n= x + iy method

Answer 19

1. Put the RHS in mod-arg form
2. Write each θ as (θ + 2kπ)
3. Use de Moivre’s theorem to raise each side by 1/n
4. Substitute k=0, k=1… for n values of k and put in principal argument form

Question 20

Geometric problems in complex roots of unity.

Answer 20

No matter what the roots are the ratio between them is the same as the roots of unity for that power
ω = cos(2π/n) + isin(2π/n)
Multiply by ω to get as many points as necessary, use exponential form

Question 21

Series of cos and sin from an infinite series

Answer 21

Find the real and imaginary parts of the series

Question 22

Infinite series from (cos θ + cos 2θ + cos 3θ) + i(sin θ + sin 2θ + sin 3θ)

Answer 22

Write as z + z^2 + z^3 + …
Where z = e^iθ
A 1 first can be z^0

Question 23

Infinite series from (cos θ + kcos 2θ + k^2cos 3θ) + i(sin θ + ksin 2θ + k^2sin 3θ)

Answer 23

Write as regular sum of series with e^iθ as the numerator
Multiply by the denominator with the power of e inversed
Write in mod arg and simplify with real and imaginary

Question 24

What is the general way nth terms of unity can be expressed as:

Answer 24

Question 25

What is the sum of all complex roots of unity?

Answer 25

This would be zero. If we take all the complex roots of unity as vectors, and connect them from ‘nose to tail’ then it will return back to the origin.

Question 26

What does the complex roots of unity of z^3 -1 =0 look like?

Answer 26

Question 27

If we are finding the complex roots of unity of (z-a)^(n) -1 = 0, what does this look like geometrically:

Answer 27

This forms a circle of complex roots of unity centred at (a, 0) with the normal arrangement of the complex roots of unity like it was centred at the origin

Question 28

How to find the complex roots of unity if we have z^n = a +bi

Answer 28

Express both sides in exponential form. Then, solving for z with nθ = arctan(a/b) + 2πk, where k = 0, 1, 2… n. Then check if k results in θ still being in the domain or not.

Question 29

What is the angle around the centre of the roots of unity for the nth rooths of unity between two roots?

Answer 29

2π/n