Calculate $\int_{\mid z\mid=1} \mid z-1\mid \mid dz\mid$

Question

I am studying about complex analysis, and our class just finished Cauchy's Integral Formula, Reflection Principle, and the Residue Formula.

Now, I am trying to calculate the following complex integral:

$\int_{\mid z\mid=1}$$\mid$$z-1$$\mid$$\mid$$dz$$\mid$, where $\mid$$dz$$\mid$$=$$\mid$$z^{'}(t)$$\mid$ $dt$

Here is what I have got so far:

We can parametrize $z=e^{i\theta}$, where $0\leq \theta \leq 2\pi$ ($r=1$ since $\mid$$z$$\mid$$=1$)

Then, $\frac{dz}{d\theta}=ie^{i\theta}$, and thus $\mid$$\frac{dz}{d\theta}$$\mid$$=$$\mid$$ie^{i\theta}$$\mid$$=1$.

Thus, $\int_{\mid z\mid=1}$$\mid$$z-1$$\mid$$\mid$$dz$$\mid$$=$$\int_{0}^{2\pi}$$\mid$$e^{i\theta}-1$$\mid$$d\theta$.

Then, I don't know what to do next, since I don't know how to deal with the absolute value. Also, I don't think my way of doing this is right, since it looks a little bit weird..

Any hints or detailed explanations are really appreciated!!!

Answer 1

HINT

Let use $$e^{ix}= \cos x+i \sin x$$

and then

$$|e^{i\theta}-1|=(2-2\cos \theta)^\frac12=2\left|\sin \frac{\theta}{2}\right|=2\sin \frac{\theta}{2}$$