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The following is an exercise from Complex Analysis by Stephen Fisher.

Fix a complex number $a$ and a positive real number $R$. Suppose $u$ is a function defined on the circle of radius $R$ centered at $a$. Let $C$ denote this circle.

Show that the average value of $u$ on $C$ is given by $\frac{1}{2\pi}\int_{0}^{2\pi} u(a + Re^{it})dt$.

Any Hints please.

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What is the definition of the average value? (The integral is usually used as the definition, but I don't know the book you are using.) –  Lukas Geyer Oct 26 '12 at 6:31
    
The exercise is probably meant to relate to Cauchy's integral formula, but as said above: the integral you write is somewhat the definition of "average value" and the goal of the calculation, so the exercise is a bit unclear in its current form. I have the book (2nd edition at least) - which exercise number/page is it, so I can get a context? –  Daniel Andersson Oct 26 '12 at 10:29
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Your accept rate is waaaaay below average: don't you like the answers you get here? Enhancing it is a good idea in order to have a higher probability to get more answers... –  DonAntonio Oct 27 '12 at 1:21
    
I think you will get more help if you improve on the 14% accept rate. See this. –  JohnD Jan 12 '13 at 16:15

1 Answer 1

We assume Cauchy's integral theorem.

$$(1)\hspace{5mm}\frac{1}{2\pi}\int_{0}^{2\pi} u(a+ Re^{i t})dt = \frac{1}{2\pi i}\int_{0}^{2\pi}\frac{u(a+Re^{it})iRe^{it}}{Re^{it}}dt $$

The right side of (1), letting R be the radius of circle $C,$ $u(z)$ the equation of the circle $|z-a|= R $ or $z = a + Re^{it},$ so that $z-a = Re^{it}$ and $dz = iRe^{it},$ is precisely

$$(2)\hspace{5mm}\frac{1}{2\pi i}\oint_C\frac{u(z)}{z-a}dz$$

By Cauchy's integral formula (2) is equal to $u(a).$ $\square $

This exercise is odd because the starting integral is generally taken as the definition of the average value, as noted in a comment above. We would normally derive that form for the average from Cauchy's integral formula.

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