Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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
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
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.