Conjugate of e^(iz)

[guildmage]

1. The problem statement, all variables and given/known data

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.

2. Relevant equations



3. 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:wink:.....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