Is this complex integral well solved?

Question

I have this exercise:

$$\int_{|z|=1}\frac{\cos z}{z^3}dz$$

The way I tried to solve it was:

Since we have a singularity in $0$, and it is inside of the curve, lets consider the new curve: $|z|=1/2$, then by the deformation theorem and Cauchy's integral formulas, we get that: $$\int_{|z|=1}\frac{\cos z}{z^3}dz=\int_{|z|=\frac 12}\frac{\cos z}{z^3}dz=2\pi i(-\cos (0))=-2\pi i$$

Is this correct? I think I followed all the hypothesis correctly, but this puzzles me, since $0$ is still inside the circle.

Answer 1

Theorem.$^1$ If $f$ is analytic inside and on the simple closed positively oriented contour $\Gamma$ and if $z$ is any point inside $\Gamma$, then $$ f^{(n)}(z)=\frac{n!}{2\pi i}\int_{\Gamma}\frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta\quad(n=1,2,3,\dots). $$

By the theorem stated above, we get $$ \int_{|z|=1}\frac{\cos z}{z^3}dz = \frac{2\pi i}{2!}(-\cos 0)=-\pi i. $$ Also, the deformation theorem is not needed as Daniel Fischer commented.

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$^1$ Fundamentals of Complex Analysis: with Applications to Engineering and Science, page 211