# Contour Integral around the unit circle $C$: $\oint_C \frac{e^z-1}{\sin^3(z)}dz$

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!

• 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 $z=0$. Jul 25, 2023 at 1:50

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

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

• 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! Jul 25, 2023 at 3:30

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 $$\frac{d}{dz}$$.

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