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..
 A: First, we have
$$\overline{\lim_{N\to \infty}S_N}=\lim_{N\to \infty}\overline{S_N}$$
Second, we have
$$\overline{S_N}=\overline{\sum_{k=0}^Nf_k(z)}=\sum_{k=0}^N\overline{f_k(z)}$$
Third, we have
$$\overline{f_k(z)}=\overline{\left(\frac{z^k}{k!}\right)}=\frac{\overline{z^k}}{k!}$$
Finally, we have
$$\overline{z^k}=\bar z^k$$
and we are done, as this last equality follows inductively from the fact that $\overline{z_1z_2}=\bar z_1\bar z_2$!  
A: 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.
A: We know that 
$$\sum_{k=0}^\infty\frac{z^k}{k!}$$
converges for all $z\in\Bbb C$.  Let's let $a_k=1/k!$, and $f(z)=e^z$.
Now,
$$\begin{align}\overline{f(z)} & =\overline{\sum_{k=0}^\infty a_kz^k}\\
& = \sum_{k=0}^\infty a_k\overline z^k,\quad\text{because }a_k\in\Bbb R\\
& = f(\overline z).
\end{align}$$
A: Let $x^*$ denote the complex  conjugate of $ x$ ) .We have $$e^z= x_n(z)+\sum_{k=0}^{k=n} z^n/n!$$  and $$e^{z^*}=x_n(z^*)+\sum_{k=0}^{k=n}(z^*)^n/n!$$  where $(x_n(z))_{n \in N}$  and $(x_n(z^*))_{n \in N}$ are sequences  converging to $0$  . Therefore $$(e^z)^*-e^{z^*}=(x_n(z))^*-x_n(z^*)$$.In the above equation the LHS is independent of $n$ and the  RHS converges to $0$ as $n \to \infty$. Therefore the LHS is $0$.This method will apply for any convergent power series whose co-efficients are all real numbers.
