Evaluating contour integral of complex conjugate

Question

This is part of a homework assignment. Any hints will be useful, I haven't made any progress. I need to evaluate: $$ \int_{\left\vert z - 1\right\vert = 1} \overline{z}^{\,n}\,\mathrm{d}z\,,\quad n \in \mathbb{Z} $$

Answer 1

Let $z=1+e^{i t}$, where $t \in [0,2 \pi]$. Then the integral is equal to

$$i \int_0^{2 \pi} dt \, e^{i t} \, \left ( 1+e^{-i t} \right )^n $$

If you expand the binomial using the binomial theorem, you will find that only one of the terms results in a non-zero integral. From there, getting to the final answer is simple. The result is $i 2 \pi n$. Note that this result is at odds with the result predicted by Cauchy's theorem, which is zero. Why?

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\left.\int_{\verts{z - 1}\ =\ 1}\,\,\, \overline{z}^{\,n}\,\dd z \,\right\vert_{\,n\ \in\ \mathbb{Z}}}:\ {\Large ?}}$.

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Note that $\ds{1 =\pars{z - 1}\pars{\overline{z} - 1} = \pars{z - 1}\overline{z} - z + 1 \implies \overline{z} = {z \over z - 1}}$.

Then, \begin{align} &\bbox[5px,#ffd]{\left.\int_{\verts{z - 1}\ =\ 1}\,\,\, \overline{z}^{\,n}\,\dd z \,\right\vert_{\,n\ \in\ \mathbb{Z}}} = \int_{\verts{z - 1}\ =\ 1}\,\,\,{z^{n} \over \pars{z - 1}^{n}}\,\dd z \\[5mm] = &\ \int_{\verts{z}\ =\ 1}\,\, {\pars{1 + z}^{n} \over z^{n}}\,\dd z \end{align}

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• $\ds{\large n < 0:}$ The integrand is singular at the "contour point" $\ds{z = -1}$.

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• $\ds{\large n = 0:}$ The integral vanishes out: There's not any pole inside the contour.

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• $\ds{\large n > 0:}$ \begin{align} &\bbox[5px,#ffd]{\left.\int_{\verts{z - 1}\ =\ 1}\,\,\, \overline{z}^{\,n}\,\dd z \,\right\vert_{\,n\ \in\ \mathbb{Z}_{\,\geq\ 1}}} = \int_{\verts{z}\ =\ 1}\,\,\,{\pars{1 + z}^{n} \over z^{n}} \,\dd z \\[5mm] = &\ 2\pi\ic{n \choose n - 1} = \bbx{2n\pi\ic} \\ & \end{align}