Conjugate e^(iz): Solving for z = nπ

[guildmage]

Homework Statement

I want to show that

$$\exp \left( i\bar{z} \right) = \overline{\exp \left( iz \right)}$$

if and only if

$$z = n\pi$$

for any integer n.

Homework Equations

The Attempt at a Solution

Utilizing Euler's formula, I got

$$\cos \bar{z} = \cos z$$

and

$$\sin \bar{z} = -\sin z$$

Though not fully convinced, I concluded that

$$\bar{z} = z$$

This then led me to

$$\sin z = 0$$

This obviously led me to the needed conclusion. But was I correct?

 

[Office_Shredder]

Euler's formula is only valid for z a real number. Taking conjugates doesn't get you very far there. You should split z into x+iy and divvy up the exponential first

 

[guildmage]

But Euler's formula has already been proven to apply even for complex numbers. My primary concern is whether or not

$$\exp \left( i\bar{z} \right) = \overline{\exp \left( iz \right)}$$

would give me

$$\bar{z} = z$$

 

[Office_Shredder]

I didn't mean valid there, I meant useful.

For example:

$$\cos \bar{z} = \cos z$$

and

$$\sin \bar{z} = - \sin z$$

is satisfied by any imaginary number (just by using the even/oddness of cosine/sine).

The problem though, is that if z is a complex number, you don't have that cos(z) and sin(z) are real numbers, so you can't draw the two relations you have

 

[lurflurf]

exp(i z)*=exp((i z)*)=exp(i*z*)=exp(-i z*)
it is also easy to see
exp(i z)=exp(-Im[z]+i Re[z])=exp(-Im[z])exp(i Re[z])

 

[gabbagabbahey]

$$\cos \bar{z} = \cos z$$

and

$$\sin \bar{z} = -\sin z$$

Though not fully convinced, I concluded that
$$\bar{z} = z$$

That's a strange conclusion...If I told you $\cos(0)=\cos(100\pi)$, would you then conclude $0=100\pi$?

 

[guildmage]

Alright. What if I say

$$\cos \bar{z} = \cos z$$

would give me

$$\bar{z} = z + 2n\pi$$

(Is this correct?)

I will then use it to say that

$$\sin \bar{z} = -\sin z$$

gives me

$$\sin \left( z + 2n\pi \right) = -\sin z$$

This implies that

$$\sin z = -\sin z$$

Therefore,

$$z = n\pi$$

 

[lurflurf]

Why even use sine?
exp(i z*)=exp(-i z*)
exp(2i z*)=1
z=n pi