# Computing $\int_{|z|=1}|z-1|\cdot|dz|$.

I'm trying to teach myself some complex integration. I'm doing some exercises, and want to compute $$\int_{|z|=1}|z-1|\cdot |dz|.$$ I parametrize with $z(t)=e^{it}$ on the interval $(0,2\pi)$, so $z'(t)=ie^{it}$, so $|z'(t)|=1$, and I believe the integral is then equivalent to $$\int_0^{2\pi}|e^{it}-1|dt.$$ How can I deal with that pesky absolute value sign to finish the computation? Thank you.

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For every real number $t$, $|\mathrm e^{\mathrm it}-1|=2|\sin(t/2)|$. –  Did Feb 13 '12 at 9:09

For the integrand in your second expression, you want to use $$|e^{it}-1|=|(\cos(t)+i\sin(t))-1|=$$ $$=\sqrt{(\cos(t)-1)^2+\sin^2(t)}=\sqrt{\cos^2(t)-2\cos(t)+1+\sin^2(t)}=$$ $$=\sqrt{2(1-\cos(t))}=\sqrt{2(1-\cos(t))}=\sqrt{4\ \sin^2\left(\frac{t}{2}\right)}=2\sin\left(\frac{t}{2}\right),$$ in your interval.

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@anon: >in your interval. –  NikolajK Feb 13 '12 at 9:43
Whoops nevermind. –  anon Feb 13 '12 at 9:55
Thank you Nick. –  Dedede Feb 13 '12 at 21:55
@Nick: As the other answers indicate, you are missing an absolute value sign (as $\sqrt{x^2} = |x|$). –  JavaMan Feb 14 '12 at 0:10
@JavaMan: Well, as I already said to username anon, I intended to make a statement for the relevant integral $[0,2\pi]$, for which the $\sin$-expression is positive. But if it bothers you, please go ahead and change it. –  NikolajK Feb 14 '12 at 8:52

Using the half-angle formula for sine, which can be easily derived from the double angle formula for cosine (for example at the end of this post), we have: $$\begin{array}{} |e^{it}-1| &=&|\cos{t}+i\sin{t}-1|\\ &=&\sqrt{(\cos{t}-1)^2+\sin^2{t}}\\ &=&\sqrt{1+\cos^2{t}+\sin^2{t}-2\cos{t}}\\ &=&\sqrt{2(1-\cos{t})}\\ &=&\sqrt{4\frac{1-\cos{t}}{2}}\\ &=&2\sqrt{\sin^2{\frac{t}{2}}}\\ &=&2\left|\sin\frac{t}{2}\right| \end{array}$$ as Didier suggested. Then use $2\int_0^{\pi}2\sin\frac{t}2dt=8\int_0^{\pi/2}\sin{u}~du$ (with $u=\frac{t}2$ and $du=\frac12dt$) to get $8$.

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Thank you bgins. –  Dedede Feb 13 '12 at 21:55
@Dedede: you're welcome, but I forgot to unstretch the $t$ axis when integrating (giving another factor of two!). I think it's correct now. –  bgins Feb 13 '12 at 23:50