Evaluate $\int_{|z|=4}\tan z dz$

Question

Evaluate $\int_{|z|=4}\tan z\,\mathrm dz$

$\tan z=\frac{\sin z}{\cos z}$ there is a a simple pole at $z=\frac{\pi}{2}$

$\operatorname{Res}(f,\frac{\pi}{2})=\lim_{z\to \frac{\pi}{2}}(z-\frac{\pi}{2})(\frac{\sin z}{\cos z})$

$\operatorname{Res}(f,\frac{3\pi}{2})=\lim_{z\to \frac{3\pi}{2}}(z-\frac{3\pi}{2})(\frac{\sin z}{\cos z})$

How to continue ?

Answer 1

You have\begin{align}\operatorname{Res}\left(\tan z,\pm\frac\pi2\right)&=\lim_{z\to\pm\frac\pi2}\left(z\mp\frac\pi2\right)\frac{\sin z}{\cos z}\\&=\lim_{z\to\pm\frac\pi2}\frac{\sin z}{\frac{\cos(z)-\cos\left(\pm\frac\pi2\right)}{z\mp\frac\pi2}}\\&=\frac{\sin\left(\pm\frac\pi2\right)}{\cos'\left(\pm\frac\pi2\right)}\\&=-1.\end{align}Besides, $\pm\frac\pi2$ are the only poles in the disc centered at $0$ with radius $4$. Therefore\begin{align}\int_{|z|=4}\tan z\,\mathrm dz&=2\pi i\left(\operatorname{Res}\left(\tan z,\frac\pi2\right)+\operatorname{Res}\left(\tan z,-\frac\pi2\right)\right)\\&=-4\pi i.\end{align}

Answer 2

Taylor expand $\cos(z)=\cos(\frac\pi2)+(z-\frac\pi 2)\cos'(\frac\pi 2) + \frac12(z-\frac\pi 2)^2 \cos''(\frac\pi 2) + \dots$ and $\cos(z)=\cos(\frac{3\pi}2)+(z-\frac{3\pi} 2)\cos'(\frac{3\pi} 2) + \frac12(z-\frac{3\pi} 2)^2 \cos''(\frac{3\pi} 2) + \dots$, use that to compute the limits, then use the residue theorem.