Solve $\int_{|z| = 3} \tan (\pi z) dz$ using argument principle

Question

$\int_{|z| = 3} \tan (\pi z) dz = \int \frac{d(\cos \pi z)}{ (-\pi)\cos \pi z} dz = -2i(N-P)$

where N = num of zeroes inside C:|z| = 3 and P is num of poles inside C (Is this correct or should we also consider on C???)

zeroes for $\cos \pi z$ = -0.5,-1.5,-2.5,0.5,1.5,2.5 $\implies $ N = 6

No poles for $\cos \pi z$

value of given integral = $-2i (N-P) = -2i(6-0) = -12i$

Is this correct? pls correct me if i am doing wrong

Answer 1

It is correct, yes. And the zeros and polse are inside $C$. Actually, if there were zeros or poles on $C$, the expressin $\int_C\frac{f'(z)}{f(z)}\,\mathrm dz$ would make no sense.