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Studying once again for my last attempt at the complex analysis qualifying exam. I'm a bit confused as to what to do with this contour integral, where $C$ is the unit circle.

$$\oint_C \frac{e^z-1}{\sin^3(z)}dz$$

I know that if we can expand into a Laurent series about $z=0$, we could get the residue there. Wolfram gives a few terms of the Laurent series including the $z^{-1}$ term which has the coefficient $1/2$. Thus, I believe the answer will be $2\pi i*1/2 = \pi i$ by the Residue Theorem since I believe that zero is the only pole inside the unit circle?

I have no idea how to get the Laurent series from this integrand though. I know that

$$e^z - 1 = z + \frac{z^2}{2} + \frac{z^3}{3!} + \frac{z^4}{4!} + \cdots $$

but expanding out $\sin^3(z)$ looks super messy and then dividing by it looks even messier. I looked at instead using

$$\sin(z) = \frac{e^{iz}-e^{-iz}}{2i}$$

but I'm still not seeing how this helps me. I also considered working with some variation of the integral

$$\oint_C \frac{e^z-1}{e^{iz}}dz$$

and then taking the imaginary part... but I'm not quite sure how to equate this with $\sin^3(z)$ (as opposed to $\sin(3z)$).

How does one work with an integral like this? Please and thank you!

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    $\begingroup$ the integral can be expressed as $\oint_C\frac{f(z)}{z^3}\,dz$ where $f(z)=\frac{z^3}{\sin ^3 z}(e^z-1)$. Notice that $f$ is analytic around $0$ and recall the general integral expression for the terms of the Laurent series of a function which is holomorphic on an annulus around $z=0$. $\endgroup$
    – Mittens
    Jul 25, 2023 at 1:50

2 Answers 2

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Surely, you can find the Laurent series, especially for the coefficient of the $-1$ power term. Since you have found

$e^z - 1 = z + \frac{z^2}{2} + \frac{z^3}{3!} + \frac{z^4}{4!} + \cdots $

For the $\sin^{-3} (z)$ term, when $z$ is near $0$, we have

$$\frac1{\sin^3z}=\frac1{(z+\mathcal O(z^3))^3}=\frac1{z^3}\cdot\frac1{(1+\mathcal O(z^2))^3}=\frac1{z^3}\cdot\frac1{1+ \mathcal O(z^2)}=\frac1{z^3}\cdot(1+\mathcal O(z^2))$$

hence

$$\frac{e^z - 1}{\sin^3z}=\left(z+\color{red}{\frac12z^2}+\mathcal O(z^3) \right)\cdot \color{red}{\frac1{z^3}}\cdot(1+\mathcal O(z^2))$$

The coefficient of the $-1$ power term is

$$a_{-1}=\frac12$$

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  • $\begingroup$ I’m not so confident in working with “order of” approximations… This works and makes sense. I need to get more comfortable with it. I confess I was kind of hoping there was a nice clean way that I was missing… Thanks! $\endgroup$
    – Serafina
    Jul 25, 2023 at 3:30
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Here is an alternate approach if familiarity with Big-$\mathcal{O}$ is not distinctive. It is a bit more cumbersome but manageable. In fact we do some kind of mimicking the usage of Big-$\mathcal{O}$. From \begin{align*} e^z-1&=z+\frac{1}{2}z^2+\frac{1}{6}z^3+\cdots\\ \sin(z)^3&=\left(z-\frac{1}{6}z^3+\cdots\right)^3 \end{align*} we see the function $f(z)=\frac{e^z-1}{\sin(z)^3}$ has a pole of order $2$ at $z=0$. We can therefore calculate the residue of $f$ at $z=0$ using the formula \begin{align*} \mathrm{res}_{z=0}f(z)=\lim_{z\to 0}\frac{d}{dz}\left(z^2f(z)\right)\tag{1} \end{align*}

We obtain \begin{align*} \color{blue}{\mathrm{res}_{z=0}f(z)}&=\lim_{z\to 0}\frac{d}{dz}\left(z^2\frac{e^z-1}{\sin(z)^3}\right)\tag{2}\\ &=\lim_{z\to 0}\frac{d}{dz}\left(z^2\frac{z+\frac{1}{2}z^2+g(z)}{(z-h(z))^3}\right)\tag{3}\\ &=\lim_{z\to 0}\frac{d}{dz}\left(\frac{1+\frac{1}{2}z+g(z)/z}{(1-h(z)/z)^3}\right)\\ &=\lim_{z\to 0}\frac{d}{dz}\left(1+\frac{1}{2}z+g(z)/z\right)\sum_{n=0}^{\infty}\binom{-3}{n}(-h(z)/z)^n\tag{4}\\ &=\lim_{z\to 0}\frac{d}{dz}\left(1\color{blue}{+\frac{1}{2}z}+g(z)/z\right)\tag{5}\\ &\,\,\color{blue}{=\frac{1}{2}} \end{align*}

Comment:

  • In (2) we calculate the residue using (1).

  • In (3) we use the series expansion of $e^z$ and $\sin(z)$ at $z=0$. We introduce $g(z)$ and $h(z)$, which are power series of order $3$. Here order is used as the smallest exponent of a term with non-negative coefficient of a power series. Note, that \begin{align*} \lim_{z\to 0}\frac{d}{dz}\left(\frac{1}{z}g(z)\right)=\lim_{z\to 0}\frac{d}{dz}\left(\frac{1}{z}h(z)\right)=0 \end{align*}

  • In (4) we make a binomial series expansion and note that $(h(z)/z)^n$ has order $2n$. So, only the constant term of the series can contribute. All other terms evaluate to $0$.

  • In (5) we see that $\frac{1}{2}z$ is the only term which is constant when applying the differentiation operator $\frac{d}{dz}$.

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  • $\begingroup$ How is $z=0$ a pole of order $3$? According to WolframAlpha here, the Laurent expansion at $z=0$ seems to be $\frac{1}{z^{2}}+\frac{1}{2z}+\frac{2}{3}+\frac{7z}{24}+\frac{7z^{2}}{30}+\mathcal{O}\left(z^{3}\right)$. $\endgroup$ Jul 31, 2023 at 0:14
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    $\begingroup$ @Accelerator: Many thanks for the hint. Answer revised. $\endgroup$ Jul 31, 2023 at 19:17

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