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So I have to evaluate $\oint \frac{z^{*}}{z-1}$ on a circle of radius 5 centered at the origin of the complex plane, where $z^{*}$ is the complex conjugate of $z$ and the orientation is anticlockwise.

I know z* isn't analytic anywhere and I can assume $z=5e^{i\theta} \implies dz = 5ie^{i\theta}d\theta.$

This transforms my integral to $\int_{0}^{2\pi}\frac{5e^{-i\theta}(5ie^{i\theta})} {5e^{i\theta}-1}d\theta = 25i\int_{0}^{2\pi}\frac{1}{5e^{i\theta}-1}d\theta$

I let $ u = 5e^{i\theta} => du = i5e^{i\theta}d\theta = iu d\theta $

$\implies \frac{25i}{i} \int \frac{1}{(u-1)u}du = 25 (\int \frac{1}{u-1}du - \int \frac{1}{u}du)$

$ = 25(ln(u-1)-ln(u)) = 25 ln (\frac{u-1}{u}) $

$\implies 25 [ln (\frac{5e^{i\theta}-1}{5e^{i\theta}})]_{0}^{2\pi}$

I'm not sure if I've made a mistake uptil now and what exactly I need to do further.

Syed Bukhari
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3 Answers3

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \oint_{\verts{z}\ =\ 5}{\overline{z} \over z - 1}\,\dd z & = \oint_{\verts{z}\ =\ 5}{\verts{z}^{2} \over z\pars{z - 1}}\,\dd z = 25\oint_{\verts{z}\ =\ 5}{\dd z \over z\pars{z - 1}} \\[5mm] & = 25\lim_{R \to \infty}\oint_{\verts{z}\ =\ R\ >\ 1} {\dd z \over z\pars{z - 1}} = \bbx{\Large 0} \\[5mm] & \mbox{because}\quad 0 < \verts{\oint_{\verts{z}\ =\ R} {\dd z \over z\pars{z - 1}}} < {2\pi \over R} \end{align}

Felix Marin
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Observe that

$$\oint \frac{z^{*}}{z-1} dz= 25i\int_{0}^{2\pi}\frac{1}{5e^{i\theta}-1}d\theta= 25 \oint \frac{1}{(z-1)z} dz.$$

Can you proceed ?

Fred
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  • I did it all the way till here (in my above solution your z = my u) and then broke it into two fractions. Can you pls check if it's correct till where i did it or if i went somewhere wrong after that because i don't know how to simplify it more. – Syed Bukhari Feb 27 '20 at 07:48
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Your approach tries the substitution $z=5e^{i\theta}$, but then reverts to $z$ (getting the same intermediate result as my comment noting that on $|z|=5$, $\bar z=\frac{25}z$), where I would normally use residues.

It is best only to use logarithms in complex analysis when one is willing to define and use a branch cut for the logarithm. Otherwise, the hidden branch cut can cause confusion.

However, let's try to continue with the $z=5e^{i\theta}$ approach. $$ \begin{align} \oint\frac{\color{#C00}{\bar z}}{\color{#090}{z-1}}\,\color{#00F}{\mathrm{d}z} &=\int_0^{2\pi}\frac{\color{#C00}{5e^{-i\theta}}\,\color{#00F}{5ie^{i\theta}}}{\color{#090}{5e^{i\theta}-1}}\,\color{#00F}{\mathrm{d}\theta}\tag1\\ &=5i\int_0^{2\pi}\frac{e^{-i\theta}}{1-\frac15e^{-i\theta}}\,\mathrm{d}\theta\tag2\\ &=5i\int_0^{2\pi}\sum_{k=0}^\infty\frac{e^{-i(k+1)\theta}}{5^k}\,\mathrm{d}\theta\tag3\\ &=5i\sum_{k=0}^\infty\frac1{5^k}\int_0^{2\pi}e^{-i(k+1)\theta}\,\mathrm{d}\theta\tag4 \end{align} $$ Explanation:
$(1)$: $z=5e^{i\theta}$
$(2)$: cancel common factors
$(3)$: use the Taylor series for $\frac1{1-x}$
$(4)$: swap order of sum and integral

Each of the integrals in $(4)$ is pretty easy.

robjohn
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