# If f is analytic on the closed disc

1. Jun 7, 2014

### DotKite

1. The problem statement, all variables and given/known data
If f is analytic on the closed disc, show that for r<1 we have

$f(re^{i\phi}) = \frac{1}{2\pi} \int_{0}^{2\pi}\frac{f(e^{i\theta})}{1-re^{i(\phi - \theta)}}d\theta$

2. Relevant equations

3. The attempt at a solution

I tried using cauchy integral formula and end up with

$f(re^{i\phi}) = \frac{1}{2i\pi} \int_{0}^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta} - re^{i(\phi)}}d\theta$

i factor the denominator and get

$f(re^{i\phi}) = \frac{1}{2i\pi} \int_{0}^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta}(1-re^{i(\phi-\theta)})}d\theta$

Don't really know where to go from here.

2. Jun 8, 2014

### CAF123

Given $z = e^{i \theta}$, did you remember to compute the transformation of $dz$?

3. Jun 8, 2014

### DotKite

Of course! Thanks a bunch man!