Why $|dz| = -i \rho \frac{dz}{z}$?

Question

I am solving exercise $3$ p. 120 in Ahlfors' C.A.

Find $$\int_{|z| = \rho} \frac{|dz|}{|z - a|^2}$$

Hint is given: $$|dz| = -i \rho \frac{dz}{z}$$

I don't understand why this is true. Please explain.

Answer 1

Perhaps the easiest way to see it is via a parametrisation: Let $z(t) = \rho e^{it}$. Then, more or less by definition (though I'm not sure how Ahlfors does it), $$ |dz| = |z'(t)|\,dt = | \rho i e^{it} | \,dt = \rho\,dt. $$ On the other hand $$ dz = z'(t)\,dt = \rho i e^{it}\,dt = iz\,dt. $$ Combining these you get $$ |dz| = \rho\,dt = \rho\,\frac{1}{iz}\,dz = -\frac{i\rho}{z}\,dz $$

Answer 2

Because of $|z|=\rho$ and thus $\Im(\rho)=0$

You can see that the absolute values of the left and the right hand side are equal:

$$|-i\rho\frac{dz}{z}|=|dz||\frac\rho z|=|dz|$$

Now when you look at the direction in the complexplain using the $\arg$ function.

With $\arg\rho=0$ and $\arg(-i)=\frac{3\pi}{2}$ you can see that

$$\arg z=\arg(dz)+\frac{\pi}2$$ This means they are perpendicular to each other ($z=i\,dz$), which can bee seen if you try to draw a sketch.