Compute
$$\int_{|z|=1} |z-1| \, |dz|$$
I'm interested in a solution verification of this problem. I'm mostly concerned with whether I accounted for $|dz|$ properly and for whether I made any algebraic errors along the way.
Solution.
Let $z(t)=\cos t + i \sin t$ with $0 \leq t \leq 2 \pi$. Then the integral we want to calculate is
$$\int_0^{2\pi} f(z(t)) \, |z'(t)| \, dt$$
where $f(z)=|z-1|$. Note that we use $|z'(t)|$ because of the $|dz|$. Now,
$$|z'(t)| = |\sin t - i \cos t| = \sqrt{((-\cos t)^2 + \sin^2t} = 1$$
Now, we are ready:
\begin{align} \int_0^{2\pi} |\cos t + i \sin t-1| \cdot (1) \, dt &= \int_0^{2\pi} \sqrt{(\cos t - 1)^2 + \sin^2t} \, dt\\ &= \int_0^{2 \pi}\sqrt{\cos^2 t -2 \cos t +1 + \sin^2t} \, dt \\ &= \int_0^{2\pi} \sqrt{2-2\cos t } \, dt\\ &= \int_0^{2\pi} \sqrt{4 \cdot \frac{1 - \cos t}{2}} \, dt \\ &= 2 \int_{0}^{2\pi} \sin\left(\frac{t}{2}\right) \, dt \\ \end{align}
Now, let $u = \frac{t}{2}$ and so $2\, du = dt$. Then
\begin{align} &= 2 \cdot 2 \int_{\dots}^{\dots} \sin(u) \, du = 4 \left(- \cos \left( \frac{t}{2} \right)\right) \Big|_0^{2 \pi} \\ &= 4 \left[ - \cos(\pi) - (- \cos(0)) \right] \\ &= 4[-(-1)-(-(1))] = 4(1+1) = 8 \end{align}