Pure 2.1 - Complex Numbers 2

Question 1

What is the exponential form of a complex number?

Answer 1

The modulus- argument form of the complex number, 𝑧=𝑟(cos(θ)+isin(θ)) where 𝑟=|𝑧| and θ=arg(𝑧), Can be rearranged into 𝑧=𝑟e^(iθ)

Question 2

Describe how cos(θ) can be written as an infinite series of powers.

Answer 2

cos(θ)=Σⁿ⁼∞ᵣ₌₀(((-1)ʳθ²ʳ)/(2r)!)

Question 3

How can eˣ be written as a series expansion in powers of x, where x∈ℝ?
How can this be applied to e^(iθ)?

Answer 3

eˣ=1+x+(x²/2!)+(x³/3!)+…+(xʳ/r!)
e^(iθ)=1+iθ+((iθ)²/2!)+((iθ)³/3!)…
=1+iθ+(i²θ²/2!)+(i³θ³/3!)+…
=1+iθ-(θ²/2!)-(iθ³/3!)+(θ⁴/4!)+(iθ⁵/5!)+…
=(1-((θ²/2!)+(θ⁴/4!)+…)+i(θ-(θ³/3!)+(θ⁵/5!)+…)
e^(iθ)=cos(θ)+isin(θ), This is known as Euler’s relation.

Question 4

Describe how to multiply and divide complex numbers in exponential form.

Answer 4

If 𝑧₁=𝑟₁e^(iθ₁) and 𝑧₂=𝑟₂e^(iθ₂), then:
𝑧₁𝑧₂=𝑟₁𝑟₂e^(i(θ₁+θ₂))
𝑧₁/𝑧₂=(𝑟₁/𝑟₂)e^(i(θ₁-θ₂))

Question 5

Describe de Moivre’s theorem.

Answer 5

For any integer 𝑛,
(𝑟(cos(θ)+isin(θ)))ⁿ=𝑟ⁿ(cos(𝑛θ)+isin(𝑛θ)
This can be proven quickly using Euler's relation;
(𝑟(cos(θ)+isin(θ)))ⁿ=(𝑟e^(iθ))ⁿ
=𝑟ⁿe^(i𝑛θ)
=rⁿ(cos(𝑛θ)+isin(𝑛θ)

Question 6

How would you prove de Moivre’s theorem for positive integer exponents?

Answer 6

1) Basis step:
𝑛=1;
LHS=(𝑟(cos(θ)+isin(θ)))¹=𝑟(cos(θ)+isin(θ))
RHS=𝑟¹(cos(1θ)+isin(1θ)=𝑟(cos(θ)+isin(θ))
2) Assumption step:
Assume de Moivre’s theorem true for 𝑛=𝑘, 𝑘∈ℤ⁺;
(𝑟(cos(θ)+isin(θ)))ᵏ=𝑟ᵏ(cos(𝑘θ)+isin(𝑘θ))
3) Inductive step
When 𝑛=𝑘+1;
(r(cos(θ)+isin(θ)))ᵏ⁺¹=(r(cos(θ)+isin(θ)))ᵏ×𝑟(cos(θ)+isin(θ))
=𝑟ᵏ(cos(𝑘θ)+isin(𝑘θ))×𝑟(cos(θ)+isin(θ))
=𝑟ᵏ⁺¹(cos(𝑘θ)+isin(𝑘θ))(cos(θ)+isin(θ))
=𝑟ᵏ⁺¹(cos(𝑘θ)cos(θ)-sin(𝑘θ)sin(θ))+i(sin(𝑘θ)cos(θ)+cos(𝑘θ)sin(θ))
=𝑟ᵏ⁺¹(cos(𝑘θ+θ)+isin(𝑘θ+θ))
=𝑟ᵏ⁺¹(cos((𝑘+1)θ)+isin((𝑘+1)θ))
4) Conclusion step
If de Moivre’s theorem is true for 𝑛=𝑘, then it has been shown to be true for 𝑛=𝑘+1.
As de Moivre’s theorem is true for 𝑛=1, it is now proven to be true for all 𝑛∈ℤ⁺.

Question 7

What trigonometric identities can be derived using de Moivre’s theorem?

Answer 7

𝑧+(1/𝑧)=2cos(θ)
𝑧ⁿ+(1/𝑧ⁿ)=2cos(𝑛θ)
𝑧-(1/𝑧)=2isin(θ)
𝑧ⁿ-(1/𝑧ⁿ)=2isin(𝑛θ)

Question 8

How would you get the sums of the series;
for 𝑤,𝑧∈ℂ,
𝑤𝑧ʳ, 0≤𝑟≤𝑛-1
𝑤𝑧ʳ, 0≤𝑟≤∞

Answer 8

Σⁿ⁻¹ᵣ₌₀𝑤𝑧ʳ=𝑤+𝑤𝑧+𝑤𝑧²+…+𝑤𝑧ⁿ⁻¹=𝑤(𝑧ⁿ-1)/(𝑧-1)

Σ(^∞)ᵣ₌₀𝑤𝑧ʳ=𝑤+𝑤𝑧+𝑤𝑧²+…=𝑤/1-𝑧, |𝑧|<1

Question 9

Describe the method of finding the 𝑛th roots of a complex number.

Answer 9

If 𝑧 and 𝑤 are complex numbers and 𝑛 is a positive integer, then the equation 𝑧ⁿ=𝑤 has 𝑛 distinct solutions.
For any complex number 𝑧=𝑟(cos(θ)+isin(θ)), you can write 𝑧=𝑟(cos(θ+2𝑘π)+isin(θ+2𝑘π)) Where 𝑘 is any integer.
Generally, the solutions to 𝑧ⁿ=1 are 𝑧=cos(2𝑘π/𝑛)+isin(2𝑘π/𝑛) = e^(2i𝑘π/𝑛) for 𝑘=1,2,3… and are known as the 𝑛th roots of unity
If 𝑛 is a positive integer, then there is an 𝑛th root of unity 𝑤=e^(2iπ/𝑛) such that:
the 𝑛th roots of unity are 1,𝑤,𝑤²,…,𝑤ⁿ⁻¹
1,𝑤,𝑤²,…,𝑤ⁿ⁻¹ form the vertices of a regular n-gon
1+𝑤+𝑤²+…+𝑤ⁿ⁻¹=0

Question 10

Describe how complex 𝑛th roots can be used to solve geometric questions

Answer 10

The 𝑛th roots of any complex number 𝑎 lie at the vertices of a regular 𝑛-gon with is centre at the origin.
If 𝑧₁ is one root of the equation 𝑧ⁿ=𝑠, and 1,𝑤,𝑤²,…,𝑤ⁿ⁻¹ are the 𝑛th roots of unity, then the roots of 𝑧ⁿ=𝑠 are given by 𝑧₁,𝑧₁𝑤,𝑧₁𝑤²,…,𝑧₁𝑤ⁿ⁻¹

Question 11

Describe how sin(θ) can be written as an infinite series of powers.

Answer 11

sin(θ)=Σⁿ⁼∞ᵣ₌₀(((-1)ʳθ²ʳ⁺¹)/(2r+1)!)