Integral of $e^{\overline{z}}$ So I'm suppose to integrate $e^{\overline{z}}$ along the unit circle where $z(t)=e^{it}$ and from t goes from 0 to $2\pi$. This is the work I've done so far
$\int e^{\overline{z}}dz = i\int_0^{2\pi}e^{e^{-it}}e^{it}dt = i\int_0^{2\pi}e^{e^{-it}+{it}}dt$
I tried solving this, getting $i[\frac{e^{it+e^{-it}}}{i+ie^{-it}}]$ from 0 to $2\pi$. Evaluating this I get 0 as the solution. To check if this is right I used wolframalpha which returned a value of $2\pi i$. Help as to where I messed up and what I should do instead would be greatly appreciated. I think the place i probably messed up is solving the integral. Thanks.
 A: $\newcommand{\+}{^{\dagger}}
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 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
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 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
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\begin{align}
\color{#00f}{\large\int_{\verts{z} = 1}\expo{\ol{z}}\,\dd z}&=
\int_{\verts{z} = 1}\expo{z\ol{z}/z}\,\dd z=\int_{\verts{z} = 1}\expo{1/z}\,\dd z
=\sum_{n = 0}^{\infty}{1 \over n!}
\overbrace{\int_{\verts{z} = 1}{1 \over z^{n}}\,\dd z}
^{\ds{=\ 2\pi\ic\,\delta_{n,1}}} = \color{#00f}{\large 2\pi\ic}
\end{align}
