Integral of $e^{\overline{z}}$

Question

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.

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}