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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f(z)$ be an entire function, $R_n$ a sequence of positive real numbers tending to $\infty$ such that $f(z) \neq 0$ on $|z|=R_n$ and there exists $M>0$ such that $$\int_{|z|=R_n} \left|\frac{f'(z)}{f(z)}\right| ~dz<M$$ for all $n$. Show that $f$ is a polynomial.

What came to my mind is to consider that $f(z)=a_0+a_1z+\cdots \;\;\forall z\in\mathbb{C}$, and to try proving that $a_k=0$ from a certain $k$, maybe using the Cauchy formula for these coefficients, but I can't use the hypotesis on that bounded integral. Is observing that there is a logarithmic derivative of any use?

share|cite|improve this question
An idea: try to show that $f^{-1}(a)$ is finite for every $a\in\mathbb C$, using the argument principle. – user31373 Jul 5 '12 at 21:55
So indicating with $N$ the number of zeros inside $|z|<R_n$ by the argument principle I have $2\pi i N=\int_{|z|=R_n}\frac{f'}{f}$ which can be estimated by its modulus and eventually by M, and taking the limit i have that there are at most M zeros. If i wanted to do something analogous with $f(z)-a$ i need an estimate like $|f(z)-a|>|f(z)|$ on $|z|=R_n$.. – balestrav Jul 5 '12 at 22:16
Right, we are stuck with $a=0$. Luckily, a better idea arrived meanwhile. – user31373 Jul 5 '12 at 22:22
up vote 8 down vote accepted

As $\,\displaystyle{\frac{1}{2\pi i}\oint \frac{f'}{f}\,dz}\,$ is the number of roots inside the circle, the total number of roots is finite

(bounded by $\,\frac{M}{2\pi}\,$). Let $$g(z)=\frac{f(z)}{(z-a_1)\cdot\ldots\cdot(z-a_k)}$$, where $a_1,\dots,a_k$ are the roots of $f$. Then $g$ satisfies the same condition as $f$ (with a different $\tilde M$). As $g$ has no root, it is of the form $g(z)=\exp(h(z))$ for some entire function $h$. We thus have $\,\displaystyle{\oint |h'(z)|\,|dz|<\tilde M}\,$ for circles of radii $R_n\to\infty$. This implies $h'=0$, hence $f$ is a polynomial.

edit: why $h'=0$: if $$h'(z)=c_1 z^m+c_2 z^{m+1}+\dots$$ ($c_1\neq0$) then $$\oint \frac{h'(z)}{z^{m+1}}\,dz=2\pi i c_1$$, which certainly implies $\oint |h'(z)|\,|dz|\to\infty$.

share|cite|improve this answer

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.