Problem with: $\int_{|z|=1} |z-1||dz|$ :contour integral with Abs. value |dz|

Question

This problem just came across in the context of a course of Complex analysis and I don't know how to tackle it. Could anyone suggest a Hint on this?

Compute $\int_{|z|=1} |z-1||dz|$ in counterclockwise direction

I really don't know what would be the meaning of |dz|, or how to treat the abs in the integrand. Any help would by highly appreciated.

Answer 1

Set $z:=e^{i\theta}$ for $0\leqslant\theta\leqslant 2\pi$ then you integral around the contour you say is equivalent to \begin{align} \int_{|z|=1}|z-1||\,dz|=\int_0^{2\pi}|e^{i\theta}-1||ie^{i\theta}\,d\theta|&=\int^{2\pi}_0\sqrt{(1-\cos\theta)^2+\sin^2\theta}\,d\theta\\&=\int^{2\pi}_0\sqrt{2-2\cos\theta}\,d\theta \end{align} Using double angle formula for the cosine $\cos2\theta=1-2\sin^2\theta$ we get \begin{align} \int^{2\pi}_0\sqrt{2-2\cos\theta}\,d\theta=\int^{2\pi}_0\sqrt{4\sin^2\theta/2}\,d\theta&=2\int^{2\pi}_0|\sin\theta/2|\,d\theta\\&=4\int^{\pi}_0\sin\theta\,d\theta=4(-\cos\theta\Big|_0^{\pi})=8 \end{align}

Answer 2

Parametrize with the usual $z=e^{it}$

$$=\int_0^{2\pi} |e^{it}-1||ie^{it}dt| = \int_0^{2\pi} \sqrt{(1-\cos t)^2+\sin^2 t} \hspace{4 pt}dt$$

From here simplify and use trig identities. (Hint: if you get $0$ you did something wrong, perhaps related to the simplification of the square root. The answer should be $8$).

Answer 3

With parametrization $$x=\cos t, y=\sin t$$ we get $$|z-1|=\sqrt {(1-\cos t}$$

Therefore the integral is $$2\int _0^{2\pi}\sin (t/2)dt =8$$