Calculate using residues $\oint\limits_\gamma\frac{\tan z}{z + 2}dz$, where $\gamma: |z + 2| = 2$ - a closed loop, run counterclockwise
Hi. Here is what I have tried:
Assuming Isolated singularity is single: $z = -2$ (equating the denominator of the integrand to $0$)
Using formula: $$ \oint_{\gamma} f(z) d z=2 \pi i \sum_{k=1}^{n} \operatorname{Res}\left(f(z) ; z=z_{k}\right) $$ we have:
$$ \oint_{|z + 2| = 2} \frac{\tan z}{z + 2} d z=2 \pi i \operatorname{Res}\left(\frac{\tan z}{z + 2}; z=-2\right)=(*) $$
$$ \operatorname{Res}\left(\frac{\tan z}{z + 2}; z=-2\right)= \lim _{z \rightarrow -2} \left[\frac{\tan z}{z + 2}\left(z-(-2)\right)\right] = -\tan 2 \approx 2,185$$
$$ (*)= 2 \pi i \cdot (\tan 2) \approx 6,2831852i\cdot(2,185) \approx 13,7288i $$
Sounds good, but I am not sure whether $z = -2$ is the only point? My teacher told me that there also $\tan z = 0$ might give some points, but she is also not sure...
If somebody knows, explain, please.
The fact that $\tan0=0$ creates no problem. However, the fact that $\tan$ is undefined at $-\frac\pi2$ does. The answer is$$2\pi i\left(\operatorname{res}_{z=-2}\frac{\tan z}{z+2}+\operatorname{res}_{z=-\pi/2}\frac{\tan z}{z+2}\right)=2\pi i\left(\tan(-2)+\frac2{\pi-4}\right).$$