Evaluate ∫|z|=4tan⁡zdz tan⁡z=sin⁡zcos⁡z there is a a simple pole at z=π2 Res(f,π2)=limz→π2(z−π2)(sin⁡zcos⁡z) Res(f,3π2)=limz→3π2(z−3π2)(sin⁡zcos⁡z) How to continue ?

You haveRes(tan⁡z,±π2)=limz→±π2(z∓π2)sin⁡zcos⁡z=limz→±π2sin⁡zcos⁡(z)−cos⁡(±π2)z∓π2=sin⁡(±π2)cos⁡′(±π2)=−1Besides, ±π2 are the only poles in the disc centered at 0 with radius 4. Therefore∫|z|=4tan⁡z,dz=2πi(Res(tan⁡z,π2)+Res(tan⁡z,−π2))=−4πi

"Evaluate ∫|z|=4tan⁡zdz tan⁡z=sin⁡zcos⁡z there is a a simple..." solved and explained

Jaylene Franco

Answered question

2022-01-29

Evaluate

\int_{| z | = 4} \tan z d z

\tan z = \frac{\sin z}{\cos z}

there is a a simple pole at

z = \frac{π}{2}

R e s (f, \frac{π}{2}) = lim_{z \to \frac{π}{2}} (z - \frac{π}{2}) (\frac{\sin z}{\cos z})

R e s (f, \frac{3 π}{2}) = lim_{z \to \frac{3 π}{2}} (z - \frac{3 π}{2}) (\frac{\sin z}{\cos z})

How to continue ?

Answer & Explanation

ma90t66690

Beginner2022-01-30Added 7 answers

You have
$R e s (\tan z, \pm \frac{π}{2}) = lim_{z \to \pm \frac{π}{2}} (z \mp \frac{π}{2}) \frac{\sin z}{\cos z}$
$= lim_{z \to \pm \frac{π}{2}} \frac{\sin z}{\frac{\cos (z) - \cos (\pm \frac{π}{2})}{z \mp \frac{π}{2}}}$
$= \frac{\sin (\pm \frac{π}{2})}{\cos^{'} (\pm \frac{π}{2})}$
$= - 1$
Besides, $\pm \frac{π}{2}$ are the only poles in the disc centered at 0 with radius 4. Therefore
$\int_{| z | = 4} \tan z, d z = 2 π i (R e s (\tan z, \frac{π}{2}) + R e s (\tan z, - \frac{π}{2}))$
$= - 4 π i$

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