Complex numbers

Question

if
Sin(θ+Φ*i)=ρ(cosα+i*sin(α))
Prove that
(i)(ρ)^2=1/2*(cosh(2Φ)-cos(2θ))

(ii)tan(α)=tanh(Φ)*cot(θ)

Answer 1

Given,

Sin(θ+Φ*i)=ρ(cosα+i*sin(α)).      <-----(A)

 

What we have to do is to take norm (absolute value) from both sides.

Remind that if z=x+iy, then |z|²=x²+y^2. If so then from (A) we get

 

|Sin(θ+Φ*i)|²=|ρ(cosα+i*sin(α))|² = ρ²---(B) 

 

(It should be clear)

 

Now in (A) to calculate |Sin(θ+Φ*i)|² we have to separate Re and Im parts of  Sin(θ+Φ*i)

 

Sin(θ+Φ*i)=Cosh(Φ) Sin(θ) + i* Cos(θ) Sinh(Φ) <

and

|Sin(θ+Φ*i)|²=(Cosh(Φ) Sin(θ))² + (Cos(θ) Sinh(Φ))^2. <---(C')

 

If you simplify the RHS expression in (C) you get 1/2*(cosh(2Φ)-cos(2θ)).

So, combining (B) and (C), we have

 

ρ² = 1/2*(cosh(2Φ)-cos(2θ)).

i) is done!

 

ii) Prove

   tan(α)=tanh(Φ)*cot(θ).

Comparing (A) and (C) we get

 

ρ cos α=Cosh(Φ) Sin(θ)

 

and  

ρ sinα = Cos(θ) Sinh(Φ)  <---(D')

 

Dividing  (D') by (D) you get

ρ sinα/ρ cos α = Cos(θ) Sinh(Φ)/Cosh(Φ) Sin(θ).

 

tan α = tan(θ) Tanh(Φ).

I didn't give all details, but they are easy to derive. So if you would have any difficulties, let me know.