Integration in the complex plane of abs(z)

Question

I have to integrate the function $|z|$ on the circle $|z-1|=1$.

I tried to parameterize the circle by $z-1=\exp(it)$, but I don't know how to proceed. Can somebody help me please? thanks.

Answer 1

As $z=1+e^{it}=1+\cos(t)+i \sin(t)$,

$$|z|=\sqrt{(1+cos(t))^2+\sin(t)^2}= \sqrt{2 + 2 \cos(t)}=\sqrt{2}\sqrt{1 + \cos(t)}=\sqrt{2}\sqrt{2 \cos(t/2)^2}=2 |cos(t/2)|$$ then your integral is

$$2\int_{t=0}^{t=2 \pi} |cos(t/2)| dt$$

Up to you for the final computation (change of variable and separating cases).