Proof that $e^{i\bar{z}}=\overline{e^{iz}}$ if and only if $z=k\pi\in Z$ I need to proof that $e^{i\bar{z}}=\overline{e^{iz}}$ if and only if $z=k\pi\in Z$, I will show you my procedure
$e^{i\overline{z}}=\overline{e^{iz}}$
$e^{i\overline{x+iy}}=\overline{e^{i(x+iy)}}$
$e^{i(x-iy)}=\overline{e^{ix-y}}$
I use $e^{x+iy}=e^x[\cos{y}-i\sin{y}]$
$e^{ix+y}=\overline{e^{-y}*[\cos{k\pi}-i\sin{k\pi}]}$
$e^{y}*[\cos{k\pi}+i\sin{k\pi}]=\overline{e^{-y}*[\cos{k\pi}+i\sin{k\pi}]}$
But $\sin{k\pi}$ always is $0$ and $\cos{k\pi}=\pm1$ if $k$ is odd $-1$ else $1$
$e^{y}*[\cos{k\pi}]=\overline{e^{-y}*[\cos{k\pi}]}$
Then if $k\pi=odd$
$e^{y}*[1]=\overline{e^{-y}*[1]}$
$e^{y}=\overline{e^{-y}}$
Else if $k\pi=pair$
$e^{y}*[-1]=\overline{e^{-y}*[-1]}$
$-e^{y}=\overline{-e^{-y}}$
but i think I am missing something... because I cant show that $e^{i\bar{z}}=\overline{e^{iz}}$
 A: Probably the easiest way: use two important facts:
(1) $\overline{e^{x+iy}}=e^{x-iy}$;
(2) the modulus of $e^{x+iy}$ is $e^x$, and the argument is $y+2k\pi$.
Then you can work as follows (the first three lines are what you have done already):
$$\displaylines{
  e^{i\overline z}=\overline{e^{iz}}\cr
  e^{i\overline{x+iy}}=\overline{e^{i(x+iy)}}\cr
  e^{i(x-iy)}=\overline{e^{ix-y}}\cr
  e^{y+ix}=e^{-y-ix}\ .\cr}$$
Now applying (2) to both sides:
$$e^y=e^{-y}\quad\hbox{and}\quad x=-x+2k\pi\ .$$
This gives $y=0$ and $x=k\pi$, so $z=k\pi$ where $k\in{\Bbb Z}$.
Don't forget that the question asked for "if and only if".  So now you have to assume $z=k\pi$, that is, $x=k\pi$ and $y=0$, and substitute in to prove that $e^{i\overline z}=\overline{e^{iz}}$.  This is pretty easy.
A: You start out strong and continue with generally the right idea, but there's at least one typo that might be causing problems. Specifically, your fourth displayed equation $e^{x + iy} = e^x[\cos y - i \sin y]$ is not always true. (You should have $\cos y + i \sin y$ here.)
Here's a suggestion: From your third displayed equality, you can conclude that
$$
e^y (\cos x + i \sin x) = e^{y+ix} = \overline{e^{ix-y}} = e^{-y} (\cos x - i\sin x)
$$
Taking absolute values allows you to conclude $e^y = e^{-y}$ and therefore (canceling factors in the other equation) $\cos x + i \sin x = \cos x - i\sin x$. For which $x$ and $y$ are these equations both true?
