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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.

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    $\begingroup$ In the context of multivariable calculus, this is the same thing as the difference between a vector and scalar line integral. In practice, just parametrize as you normally would and utilize the absolute value signs accordingly. $\endgroup$ – Ninad Munshi Oct 13 '19 at 16:05
  • $\begingroup$ Another one: math.stackexchange.com/q/2654996/42969. $\endgroup$ – Martin R Oct 13 '19 at 16:21
  • $\begingroup$ @MartinR oh I guess it is. What should I do in this case? Just typing [duplicate] in title? $\endgroup$ – holahola Oct 13 '19 at 16:21
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    $\begingroup$ No. There should be a button where you can confirm that the question is a duplicate. If not – just wait, and it will be closed by others. $\endgroup$ – Martin R Oct 13 '19 at 16:22
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    $\begingroup$ Btw, the duplicates were found with Approach0 $\endgroup$ – Martin R Oct 13 '19 at 16:24
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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}

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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 $8$).

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

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

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