Apply the Cauchy–Goursat theorem to show that $\int_{C}^{}f(z)dz=0$ when the contour C is the unit circle $\left | z \right |=1$ , in either direction, and when $f(z)=sechz$.
here is solution: Note that f is analytic on $\mathbb{C}\setminus \left \{ \frac{(2n+1)\pi }{2} \right \}$. Since $\frac{\pi }{2}>1=\left | z \right |$ , none of $\left \{ \frac{(2n+1)\pi }{2} \right \}$ lie in the region bounded by the contour $C$ which implies that $f$ is analytic on the region bounded by $C$ as well as on contour $C$. Therefore , by Cauchy-Goursat theorem,we have $$\int_{C}^{}sechz dz=0$$
Is it true that the given function analytic on $\mathbb{C}\setminus \left \{ \frac{(2n+1)\pi }{2} \right \}$? I am not sure about this solution. Any help will be appreciated.