Suppose that $f(z)$ is analytic and $f(z)\neq0$ on the disc $|z|<1$ .
Show that for $0<r<1$,
$$ \exp\left[\frac{1}{2\pi}\int_{0}^{2\pi}{\log}|f(re^{i\theta})|\,d\theta\right] = |f(0)| $$
Any other theorems needed to solve this problem? I don't have any idea just with "Gauss's MVT"
Thank you in advanced.
On
You can also use Jensen's formula : If f(z) is analytic on the disc $|z| <r $ apart from roots at a1,a2,...,am (repeated by multiplicity) and $f(0) \neq 0 $ then :
$\frac{1}{2 \pi }\int_0^{2\pi} \ln |f(re^{i\theta})|d\theta + \sum_{k=1}^{m} \ln (\frac{|a_k|}{r}) = \ln |f(0)| $
In your case, there are no zeros so you get
$\frac{1}{2\pi}\int_{0}^{2\pi}{\log}\left|f(re^{i\theta})\right|\mathrm d\theta=ln|f(0)|$
Here is a hint: you can rewrite the equality as $$\frac 1{2\pi} \int_0^{2\pi} \log |f(re^{i\theta})| \, d\theta = \log |f(0)|$$ or even just $$\frac{1}{2\pi} \int_0^{2\pi} g(re^{i\theta}) \, d\theta = g(0)$$ if you define $g(z) = \log |f(z)|$. If $f$ is analytic and nonvanishing then $g$ is harmonic.