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By finding the appropriate Laurent series, evaluate the integral where C is the unit circle, traversed once in the positive direction.

$$\int_{C} (z^3 - e^\frac{1}{z}) \cos \left(\frac{1}{z}\right)dz$$

Unsure what the Laurent Series is for the complex function.

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  • $\begingroup$ Without the 'Laurent Series', you can make the change $z \to 1/z$. $\endgroup$ Apr 7, 2018 at 3:28

2 Answers 2

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The integrand $f(z):=(z^3 - e^{1/z}) \cos (1/z)$ has "only" an essential singularity at $z=0$ (which is inside the unit circle). Hence, by the Residue Theorem, $$\int_{|z|=1} f(z)dz=2\pi i\,\mbox{Res}(f,0)=2\pi i a_{-1}$$ where $a_{-1}$ is the coefficient of $1/z$ in the Laurent expansion of $f$ in a neighbourhood of $z=0$.

In order to find $a_{-1}$ recall that $$e^{1/z}=1+\frac{1}{z}+O(1/z^2)\quad \mbox{and}\quad \cos(1/z)=1-\frac{1}{2z^2}+\frac{1}{24z^4}+O(1/z^6).$$ Can you take it from here?

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  • $\begingroup$ Use the expansions that I gave you and compute the product $(z^3 - e^{1/z}) \cos (1/z)$. What do you get? $\endgroup$
    – Robert Z
    Apr 6, 2018 at 7:31
  • $\begingroup$ You have not considered all the terms that I have written down... $\endgroup$
    – Robert Z
    Apr 6, 2018 at 7:34
  • $\begingroup$ So expanding the brackets, the only 2 terms with 1/z are (1/24z - 1/z), is this correct $\endgroup$
    – maths101
    Apr 6, 2018 at 7:41
  • $\begingroup$ hence the coefficient of 1/z is just -23/24 ? $\endgroup$
    – maths101
    Apr 6, 2018 at 7:45
  • $\begingroup$ Are you unsure about that? $\endgroup$
    – Robert Z
    Apr 6, 2018 at 7:46
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Besides any 'Laurent Expansion', you can make the change $\ds{z \mapsto 1/z}$. That yields an integrand which has a pole of order five at $\ds{z = 0}$.

Namely, \begin{align} \oint_{\verts{z}\ =\ 1}\pars{z^{3} - \expo{1/z}}\cos\pars{1 \over z}\dd z & \,\,\,\stackrel{z\ \mapsto\ 1/z}{=}\,\,\, \oint_{\verts{z}\ =\ 1} {\pars{1 - z^{3}\expo{z}}\cos\pars{z} \over z^{5}}\,\dd z \\[5mm] & = 2\pi\ic\,{1 \over 4!}\ \underbrace{\lim_{z \to 0} \totald[4]{\bracks{\pars{1 - z^{3}\expo{z}}\cos\pars{z}}}{z}}_{\ds{=\ -23}}\ =\ \bbx{-\,{23 \over 12}\,\pi\ic} \end{align}

Otherwise, you can do the following approach:

\begin{align} &{\pars{1 - z^{3}\expo{z}}\cos\pars{z} \over z^{5}} \,\,\,\stackrel{\mrm{as}\ z\ \to\ 0}{\sim}\,\,\, {\pars{1 - z^{3} - z^{4}}\pars{1 - z^{2}/2 + z^{4}/24} \over z^{5}} \\[5mm] \stackrel{\mrm{as}\ z\ \to\ 0}{\sim}\,\,\,&\ {1 \over 2} + {1 \over z^{5}} - {1 \over 2}{1 \over z^{3}} - {1 \over z^{2}} + \pars{\color{red}{-\,{23 \over 24}}}{1 \over z} \end{align}

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