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Let $r > |z|$, calculate $$\frac{1}{2 \pi} \int_{0}^{2 \pi} \ln (|r e^{it} - z|) dt. $$

My guess is that this is $\ln(r)$, obviously for $z = 0$ this goes without saying. Is there some argument for why moving the circle over which we integrate shouldn't change the outcome (as long as $z$ stays inside the disk)? I made some attempt to calculate this directly, but to no avail. Help or hints appreciated, thanks in advance.

Shaun
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TheMountainThatCodes
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1 Answers1

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Fix $r>0$

For fixed $t$ we have that $f(z)=\ln (re^{it}-z)$ is analytic in $|z|<r$ since $re^{it}-z \ne 0$ anywhere there. But now integrating a continuous map of holomorphic functions wr to a finite positive measure on a compact space leads to a holomorphic function $g$ on $|z|<r$ (with $\Re g$ being the original integral up to the $\frac{1}{2 \pi}$) since we can put the $\partial_{\bar z}$ inside and get:

$\partial_{\bar z}g(z)=\partial_{\bar z} \int_{0}^{2 \pi} \ln (r e^{it} - z) dt=\int_{0}^{2 \pi}\partial_{\bar z} \ln (r e^{it} - z) dt=0$

(or just expand in Taylor series in $z$ and show that integration and expansion commute by absolute convergence in small closed discs inside $|z| <r$).

But now differentiating we get (allowed as above) $g'(z)=-\int_{0}^{2 \pi} \frac{dt}{r e^{it} - z}=-\int_{0}^{2 \pi}\sum_{k \ge 1}\frac{z^{k-1}}{r^ke^{ikt}}dt=0$ again by switching sums and integrals by absolute convergence. So $g(z)=g(0)$ is constant and in particular:

$\frac{1}{2 \pi}\int_{0}^{2 \pi} \ln (|r e^{it} - z|) dt=\frac{1}{2 \pi}\Re g(z)=\frac{1}{2 \pi}\Re g(0)=\ln r$ so we are done!

Conrad
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