Cauchy integral formula

Question

I have the following integral

$$\int_{|z|=1}\frac {e^z}{z(z-2)} dz$$

I write, $$\int_{|z|=1}\dfrac {e^z}{z(z-2)}dz=\int_{|z|=1}\frac {Ae^z}{z}dz+\int_{|z|=1}\frac {Be^z}{z-2} dz$$

Can I now conclude that $\displaystyle\int_{|z|=1}\dfrac {Ae^z}{z}dz=0$?

How do I then compute $\displaystyle\int_{|z|=1}\frac {Be^z}{z-2}dz$?

Cauchy integral formula says that if the rectifiable path is within the domain, then the line integral is $0$. How about the integral of the point outside then? I am confused

Help would be appreciated!

Answer 1

Since $z=0$ is the only singularity of the integrand in the unit disc, you can calculate the integral as follows, where $f(z)=e^z/(z-2)$: $$\int_{|z|=1}\frac {e^z}{z(z-2)}dz=\int_{|z|=1}\frac {e^z/(z-2)}{z}dz=2\pi i\cdot f(0)=2\pi i\cdot -\frac{1}{2}=-\pi i.$$

Note: Cauchy's Integral Formula states: $$\int_{C}\frac {f(z)}{z-z_0}dz=2\pi i \cdot f(z_0),$$ where $f(z)$ is analytic in and on the simple closed contour $C$.