Is the Residue Theorem applied correctly?

Can you please tell me if I've done this exercise correctly?

Given the function $$\displaystyle g(z)=\frac{ze^{1/z}}{z^{2} + 3}\,,\,$$ find all the singularities and compute the integral $$I = \oint_{|z|=1} g(z)dz$$

First of all I found the singularities, namely $$z=0$$,$$z=\sqrt{3}I$$ and $$z=-\sqrt{3}I$$. The first on is an essential singularity while the other two are 1-poles (poles of order 1). In order to find the integral I want to apply the Residue Theorem as follows: $$I=\int_{|z|=1} g(z)dz = 2\pi iRes(g,0)$$

Since the direct computation of the Laurent series of g around z=0 seems too difficult, I tried to compute $$Res(g,0)$$ as $$Res(g,0)=-Res(g,\infty)-Res(g,\sqrt{3}i)-Res(g,-\sqrt{3}i)$$

If we set $$h(\omega)=-\frac{1}{\omega^2}g(\frac{1}{\omega})$$, we can compute $$Res(g,\infty)=Res(h,0)$$

$$-\frac{1}{\omega^2}g(\frac{1}{\omega})= -\frac{1}{\omega^2}\left(\frac{\frac{1}{\omega}e^\omega}{\frac{1}{\omega^2}+3}\right)=-\frac{1}{\omega}\frac{e^\omega}{1+3\omega^2}$$

Now if I consider a suitable area around $$\omega=0$$ I can develop $$\frac{1}{1+3\omega^2}$$ as a geometric series:

$$\frac{1}{1+3\omega^2} = \sum_{n=0}^{+\infty}(-1)^n(3\omega)^n$$

So since

$$-\frac{1}{\omega^2}g(\frac{1}{\omega})=-\frac{1}{\omega}\frac{e^\omega}{1+3\omega^2} =- \frac{1}{\omega}(1+\omega+\frac{\omega^2}{2}+\dots)(1-3\omega^2+\dots)=-\frac{1}{\omega}+\dots$$

Thus $$Res(g,\infty)=-1$$ and after some computation I found that $$I=2\pi i( 1-cos(\frac{\sqrt{3}}{3}))$$.

So, my question is: are my passages legit? Is there another way to compute $$Res(g,0$$)?

Thank you!

• $$\int_{|z|=1} g(z)dz = 2\pi iRes(g,0).$$ Commented Jun 12 at 0:43
• Oh yes of course, I forgot the $2\pi i$ Commented Jun 12 at 1:45
• You make a mistake in $\frac{1}{\omega}\frac{e^\omega}{1+3\omega^2} = \frac{1}{\omega}(1+\omega+\frac{\omega^2}{2}+\dots)(1-3\omega^2+\dots)$, the coefficient of $1/\omega$ is 1, and $Res(g,\infty)=-1$. If so, you would also get the right result. Commented Jun 12 at 1:57
• Note that $g(z)$ is not a meromorphic function since it has an essential singularity at 0 Commented Jun 12 at 15:55
• Right now I get it @Riemann thank you ! Commented Jun 12 at 20:31

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} \color{#44f}{\Large\tt I} & \equiv \color{#44f}{\oint_{\verts{z}\ =\ 1} \,\,\,{z\expo{1/z} \over z^{2} + 3}\,\dd z} \sr{z\ \mapsto\ 1/z}{=} {1 \over 3}\oint_{\verts{z}\ =\ 1} \,\,\,\overbrace{\,\,\expo{z} \over z\pars{z^{2} + 1/3}\,\,}^{\ds{\equiv \on{f}\pars{z}}}\,\ \dd z \\[5mm] & = {1 \over 3}\,2\pi\ic\left\{\rule{0pt}{7mm}% \on{Res}\bracks{\rule{0pt}{5mm}\on{f}\pars{z}, z = 0}\right. \\ & \left.\phantom{{1 \over 3}\,2\pi\ic\braces{\!\!\!}}+ \on{Res}\bracks{\rule{0pt}{4mm}\on{f}\pars{z}, z = {\root{3} \over 3}\,\ic} + \on{Res}\bracks{\rule{0pt}{4mm}\on{f}\pars{z}, z = -\,{\root{3} \over 3}\,\ic}\right\} \\[5mm] & = {2\pi\ic \over 3}\bracks{% 3 + \pars{-\,{3 \over 2}\expo{\root{3}\ic/3}} + \pars{-\,{3 \over 2}\expo{-\root{3}\ic/3}}} \\[5mm] & = 2\pi\ic\bracks{1 - \cos\pars{\root{3} \over 3}} = \bbx{\color{#44f}{4\pi\ic\,\sin^{2}\pars{\root{3} \over 6}}} \approx 1.0184\,\ic\\ & \end{align}
$$\frac{ze^\frac{1}{z}}{z^2+3} =\frac13z\cdot\frac{1}{1+\frac{z^2}{3}}\cdot e^{\frac1z}\\ =\frac13z\cdot\left(1-\frac{z^2}{3}+\frac{z^4}{3^2}-\frac{z^6}{3^3}+\cdots\right) \cdot \left(1+\frac1z+\frac{1}{2!}\frac1{z^2}+\frac{1}{3!}\frac1{z^3}+\frac{1}{4!}\frac1{z^4}+\frac{1}{5!}\frac1{z^5}+\cdots\right)\\ =\cdots+\frac13\left(\frac{1}{2!}+\frac{-1}{3\cdot4!}+\frac{1}{3^2\cdot6!}+\cdots\right)\frac1z+\cdots.$$ So the coefficient of $$\frac1z$$ is $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n)!}\frac{1}{3^n} =\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n)!}\left(\frac{1}{\sqrt3}\right)^{2n} =1-\cos{\frac1{\sqrt3}}.$$ Use $$\int_{|z|=1} g(z)dz = 2\pi i\cdot Res(g,0),$$ you can get $$\int_{|z|=1} g(z)dz=2\pi i\left(1-\cos{\frac1{\sqrt3}}\right).$$