Jaylene Franco

2022-01-29

Evaluate ${\int }_{|z|=4}\mathrm{tan}zdz$
$\mathrm{tan}z=\frac{\mathrm{sin}z}{\mathrm{cos}z}$ there is a a simple pole at $z=\frac{\pi }{2}$
$Res\left(f,\frac{\pi }{2}\right)=\underset{z\to \frac{\pi }{2}}{lim}\left(z-\frac{\pi }{2}\right)\left(\frac{\mathrm{sin}z}{\mathrm{cos}z}\right)$
$Res\left(f,\frac{3\pi }{2}\right)=\underset{z\to \frac{3\pi }{2}}{lim}\left(z-\frac{3\pi }{2}\right)\left(\frac{\mathrm{sin}z}{\mathrm{cos}z}\right)$
How to continue ?

ma90t66690

Expert

You have
$Res\left(\mathrm{tan}z,±\frac{\pi }{2}\right)=\underset{z\to ±\frac{\pi }{2}}{lim}\left(z\mp \frac{\pi }{2}\right)\frac{\mathrm{sin}z}{\mathrm{cos}z}$
$=\underset{z\to ±\frac{\pi }{2}}{lim}\frac{\mathrm{sin}z}{\frac{\mathrm{cos}\left(z\right)-\mathrm{cos}\left(±\frac{\pi }{2}\right)}{z\mp \frac{\pi }{2}}}$
$=\frac{\mathrm{sin}\left(±\frac{\pi }{2}\right)}{\mathrm{cos}{}^{\prime }\left(±\frac{\pi }{2}\right)}$
$=-1$
Besides, $±\frac{\pi }{2}$ are the only poles in the disc centered at 0 with radius 4. Therefore
${\int }_{|z|=4}\mathrm{tan}z,dz=2\pi i\left(Res\left(\mathrm{tan}z,\frac{\pi }{2}\right)+Res\left(\mathrm{tan}z,-\frac{\pi }{2}\right)\right)$
$=-4\pi i$

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