$\int_\gamma\frac{1}{|\xi-z|}\,d|\xi|=2\pi$?

Question

If $z$ is inside $\gamma$, we know $\int_\gamma\frac{1}{|\xi-z|}\,d|\xi|=2\pi$ when $\gamma$ is the circle centered at $z$, I wonder if it's still true if $\gamma$ is an arbitrary closed curve with $z$ inside?

Answer 1

Your integral is wrong.

Since $\gamma$ is a circle about $z$, $|w-z|=R$($R$ is the radius of the circle.

Thus, the integral equals $$\frac1R\int_\gamma d|w|=\frac1R(|w(2\pi)|-|w(0)|)=0$$

I think you mean $$\left\vert\int_\gamma \frac1{w-z}dw\right\vert=2\pi$$ and you wrongly assumed $$\left\vert\int_\gamma \frac1{w-z}dw\right\vert=\int_\gamma \frac1{|w-z|}d|w|$$

Added:

I might have omitted some details in the original answer.

The parametrization is: $w=z+Re^{i\theta}$. $$|w|=\sqrt{z^2+R^2+2Rz\cos(\theta)}$$ $$\int_\gamma d|w|=\int^{2\pi}_0\frac{d|w|}{d\theta}d\theta =|w|_{\theta=2\pi}-|w|_{\theta=0}=0 $$

ADDED 2:

$d|w|$ does not represent the differential length of curve, but $|dw|$ does.

This is correct: $$\int_\gamma\frac{1}{|w-z|}|dw|=2\pi$$