equality of complex analysis property I need to prove the following that
$e^{\bar{z}} = \bar{e^{z}}$ 
with $e^z :=  \Sigma_{k = 0}^{k = \infty} \frac{z^k}{k!}$
The problem that I am having is first we know $e^z$ defined that way always converge, but what I don't understand is we have to see that $\bar{z^k}$ put into the formula above converges to same number as if we do the summation and then converge to a number, and conjugate that. I don't know how to do that..

Say $z=x+iy$. Then $$e^{\bar{z}}=e^xe^{-iy}=e^x\cos(y)-ie^x\sin(y)$$ $$\bar{e^{z}}=\bar{e^x\cos(y)+ie^x\sin(y)}=e^x\cos(y)-ie^x\sin(y).$$
To do it your way: conjugation is continuous (it is an isometry), so it respects limits of sequences.
