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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2For every real number $t$, $|\mathrm e^{\mathrm it}-1|=2|\sin(t/2)|$. – Did Feb 13 '12 at 09:09
2 Answers
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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.
Nikolaj-K
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@Nick: As the other answers indicate, you are missing an absolute value sign (as $\sqrt{x^2} = |x|$). – JavaMan Feb 14 '12 at 00:10
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@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. – Nikolaj-K Feb 14 '12 at 08:52
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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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@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