Evaluate the following closed contour integral

1)

$\displaystyle \oint\frac{zdz}{z-1}$

with $\displaystyle |z|=2 $

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

Let $\displaystyle f(z) = z$

And by the Cauchy Integral formula

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

So

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

$\displaystyle z_0 = 1 $

Thus

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

Is this right?

But in this following question, I have a problem

$\displaystyle \oint\frac{zdz}{z-2}$

with $\displaystyle |z|=1$

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

How should I go upon this question?