Given the complex contour integral $\int_\alpha |z|\,|\mbox{d}z|$, with $\alpha(t)=\mbox{e}^{it}$, $0\leq t\leq 2\pi$. What does $|\mbox{d}z|$ mean? My guess is: $$\frac{|\mbox{d}z|}{|\mbox{d}t|}= \frac{|\mbox{de}^{it}|}{|\mbox{d}t|}=\left|\frac{\mbox{de}^{it}}{\mbox{d}t}\right|=|i\mbox{e}^{it}|=1 \implies |\mbox{d}z|=|\mbox{d}t|.$$ Is this correct? Thanks in advance.
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Note that $|z|=1$ on your curve, so that your integral can be written as $\int_\alpha |dz|=\int_\alpha |z'(t)|,dt$ which should give the arc-length of $\alpha$ (which is $2\pi$.) That's consistent with your conclusion above, as you can check if you plug it into the integral. – Semiclassical Aug 17 '14 at 14:54
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1Then what does $|dt|$ means ? – user10676 Aug 17 '14 at 14:55
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@user10676 See my answer below. – Fly by Night Aug 17 '14 at 14:59
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Yeah, that's right. You're making the substitution $z=\alpha(t)$ and so $\mathrm{d}z=\alpha'(t)~\mathrm{d}t=\mathrm{i}\mathrm{e}^{\mathrm{i}t}~\mathrm{d}t$.
As you say: $|\mathrm{d}z|=|\alpha'(t)~\mathrm{d}t|=|\mathrm{i}\mathrm{e}^{\mathrm{i}t}~\mathrm{d}t|=|\mathrm{i}\mathrm{e}^{\mathrm{i}t}|~|\mathrm{d}t|=1~|\mathrm{d}t|=|\mathrm{d}t|$
Since $t$ is real and $t$ increases from $0$ to $2\pi$ you also have $|\mathrm{d}t|=\mathrm{d}t$.
Had $t$ decreased from $0$ to $-2\pi$ then your would have had $|\mathrm{d}t|=-\mathrm{d}t$.
Informally $|\mathrm{d}z|$ is not interested in the direction in which $z$ is changing, but only the size of the increments of $z$. Your integral is measuring arc-length, so you don't care about where the curve goes, but only how far it goes. You don't care about $\mathrm{d}z$, but only $|\mathrm{d}z|$.
Fly by Night
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