# Math Help - Cauchy Integral #2

1. ## Cauchy Integral #2

Evaluate the following closed contour integral

1)
$\oint\frac{zdz}{z-1}$
with $|z|=2$

Since $z =1$ is inside $|z| =2$

Let $f(z) = z$

And by the Cauchy Integral formula
$\oint\frac{f(z)}{z-z_0} = 2\pi i f(z_0)$

So
$\oint\frac{z}{z-z_0} = 2\pi i f(z_0)$
$z_0 = 1$

Thus
$2\pi i * f(z_0) = 2 \pi i$

Is this right?

But in this following question, I have a problem
$\oint\frac{zdz}{z-2}$
with $|z|=1$

As $z=2$ is not inside $|z|=1$

How should I go upon this question?

2. The solution to 1) is correct. For the second problem, the function z/(z–2) is analytic everywhere inside the circle |z|=1, so by Cauchy's theorem the integral is 0.