1
$\begingroup$

Suppose that $f(z)$ is holomorphic on the closed unit disk $\bar U$ and never vanishes on the boundary $\partial \bar U$. Prove that the maximum of $\displaystyle \operatorname{Re}\frac{zf'(z)}{f(z)}\geq\#\mbox{zeros of $f(z)$ in the unit disk.}$

Thoughts: By the argument principle$\displaystyle \int_{\partial \bar U}\frac{f'(z)}{f(z)}\mathrm{d}z=2\pi i\,\#\mbox{zeros}$. There is a similarity between the pattern of $\frac{f'}{f}$ and $\frac{zf'}{f}$, plus $\displaystyle\int \frac{zf'}{f}\mathrm{d}z=2\pi i\sum\mbox{roots}$ as a consequence of the argument principle. This might give us some ways. Another idea is to use the Borel-Caratheodory inequality.

$\endgroup$
1
$\begingroup$

I assume there is a typo in you question and the real question is how to show the maximum of $\Re\left[ \frac{zf'(z)}{f(z)}\right]$ over $\partial\bar{U}$ is greater than or equal to the number of zeros over $U$. Otherwise, your question is simply false.

In any event, you don't really need anything complicated. You just need to rewrite the formula for number of zeroes as an integral over $[0,2\pi]$. $$ \begin{align} \#\text{zeroes} &= \frac{1}{2\pi i}\int_{\partial \bar{U}} \frac{f'(z)}{f(z)}dz = \frac{1}{2\pi}\int_0^{2\pi} \frac{e^{i\theta} f'(e^{i\theta})}{f(e^{i\theta})}d\theta = \frac{1}{2\pi}\int_0^{2\pi} \Re\left[\frac{e^{i\theta} f'(e^{i\theta})}{f(e^{i\theta})}\right]d\theta \\ &\le \max_{z\in\partial\bar{U}}\,\Re\left[ \frac{zf'(z)}{f(z)}\right] \end{align} $$

$\endgroup$
  • $\begingroup$ you are right there was a typo. sorry for that $\endgroup$ – Wilson of Gordon Aug 28 '14 at 3:45
  • $\begingroup$ You answer is right. Thanks. $\endgroup$ – Wilson of Gordon Aug 28 '14 at 3:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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