# Apply Cauchy-Goursat Theorem to show that $\int_C{f(z)\ dz=0}$

Apply Cauchy-Goursat Theorem to show that $$\int_C{f(z)\ dz=0}$$ when the contour $$C$$ is the unit circle $$|z|=1$$, in either direction, and when $$f(z) = \tan{z}$$

I know $$\tan(z)$$ is not entire. The problem point occurs when $$\cos(z)=0$$ at $$\frac{\pi}{2}, \ \frac{3\pi}{2}$$. Since these point occur $$\underline{\text{on}}$$ the unit circle, how do I handle that?

I know that my function has to be analytic on and interior to $$C$$, and that when the problem point occurs outside of the radius of the unit circle, Cauchy Goursat applies. But what do you do when the problem point is inside or on your curve?

• These points don't lie on the unit circle. $\lvert \pi/2\rvert > 1$. $e^{i\pi/2}$ lies on the unit circle, but $\pi/2$ doesn't. – Daniel Fischer Nov 10 '16 at 23:22

## 1 Answer

Hint. Since we know that $$\tan z=\frac{\sin z}{\cos z}$$ and that $\cos$ is analytic around $z=0$ the nearest singularities of $\tan z$ can be found at $z=\pm \pi/2$ thus there are no singularities inside the unit circle.