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In Gamelin's Complex Analysis, the expression for the average value of a complex function on a circle is introduced before Cauchy's integral theorem.

It says that the average value of $h(z)$ on the circle $|z-z_0|=R$ is given by $$ A(r) = \frac{1}{2 \pi} \int_0^{2 \pi} h\left(z_0 + re^{i \theta}\right) d \theta. $$

I'm puzzled as to why we are normalizing by $2\pi$ instead of $2\pi R$.

gt6989b
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user49404
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  • This post may help: https://math.stackexchange.com/questions/221375/average-value-of-a-complex-valued-function-on-a-circle. – JMJ Apr 20 '17 at 18:07
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    @AmitaiYuval you should take a look at this essential duplicate from a few years ago--a very elegant approach. – JMJ Apr 20 '17 at 18:13
  • @ALB I just had a look at that old question, and I must say - the answer you're talking about is terribly wrong! I wish I knew about it earlier so I could write a comment while it was relevant. To begin with, $u$ is not necessarily holomorphic, so Cauchy's integral means nothing. Second of all, Cauchy's integral gives one the value of a holomorphic $u$ at the center point. – Amitai Yuval Apr 20 '17 at 21:35
  • @ALB The interesting observation is, that combining Cauchy's theorem with the weird solution you referred me to, we see that the value of a holomorphic function at a point is equal to the average value of the function on a circle around the point. This makes sense, as both the real and imaginary parts of a holomorphic function are harmonic. – Amitai Yuval Apr 20 '17 at 21:38
  • @AmitaiYuval I am not sure the situation is as dire as you make it out to be. Certainly a meromorphic function for which every singularity is removable can also be represented via the integral formula. As for the point regarding the average, I agree with you, but I believe the writer's point (as mine) was meant to be interpreted for functions which at worst have an analytic extension. – JMJ Apr 20 '17 at 21:55
  • @ALB The subject of both posts (this one and the old one) is the average value of a general complex function. Such a function can be very far away from being holomorphic. It does not even have to be differentiable. The Cauchy formula is simply inapplicable. – Amitai Yuval Apr 20 '17 at 22:37

1 Answers1

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It is a matter of parametrization. In the integral you write, the circle is parametrized by$$\gamma:[0,2\pi]\to\mathbb{C},\quad \theta\mapsto z_0+re^{i\theta}.$$Then you divide by $2\pi$, which is the length of the interval $[0,2\pi]$.

I guess this puzzles you because this is not an arc-length parametrization. An equivalent way to calculate the average value is by$$A=\frac{1}{2\pi r}\int_0^{2\pi r}h\left(z_0+re^{it/r}\right)dt.$$This uses the parametrization$$\gamma:[0,2\pi r]\to\mathbb{C},\quad t\mapsto z_0+re^{it/r},$$which is arc-length.

Amitai Yuval
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