Computing complex integral with absolute value What is the value of $\int_{|z|=1}|z-1||dz|$?
By definition, if $z$ is parametrized by $z=e^{i\theta}$, the integral is $$\int_0^{2\pi}|e^{i\theta}-1|\cdot|ie^{i\theta}|d\theta = \int_0^{2\pi}|e^{2i\theta}-e^{i\theta}|d\theta$$ It has an absolute value, so I don't know how to integrate it.
 A: The differential $|dz|$ is an infinitely small increment of arc length.  In fact
$$
|dz| = |d(e^{i\theta})|=|ie^{i\theta}\,d\theta| = d\theta.
$$
As pointed out by njguliyev,
$$
|e^{i\theta}-1| = \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta} = \sqrt{2-2\cos\theta}.
$$
So you have
$$
\int_0^{2\pi} \sqrt{2-2\cos\theta} \, d\theta.
$$
The tangent half-angle substitution $t = \tan\dfrac\theta2$ leads via trigonometric identities to
$$
\cos\theta=\frac{1-t^2}{1+t^2}\text{ and }d\theta=\frac{2\,dt}{1+t^2},
$$
and as $\theta$ goes from $0$ to $2\pi$, then $t$ goes first from $0$ to $\infty$ and then from $-\infty$ to $0$.
The integral becomes
$$
\int_{-\infty}^\infty \sqrt{1-\frac{1-t^2}{1+t^2}} \, \frac{2\,dt}{1+t^2}
= \int_{-\infty}^\infty \frac{2\sqrt{2}|t|\,dt}{\sqrt{1+t^2}^3}
= 2\int_0^\infty\frac{2\sqrt{2}t\,dt}{\sqrt{1+t^2}^3}.
$$
The factor $|t|$ comes from the fact that $\sqrt{t^2}=|t|$ and vanishes when the bounds exclude negative values of $t$.
Then let $u=1+t^2$ so that $du=2t\,dt$ and we get
$$
2\int_1^\infty \frac{2\sqrt{2}\,du}{u^{3/2}}.
$$
Note that as $t$ goes from $0$ to $\infty$, $u$ goes from $1$ (not $0$) to $\infty$.
