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!
Best Answer
Surely, you can find the Laurent series, especially for the coefficient of the $-1$ power term. Since you have found
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
$$a_{-1}=\frac12$$