TITLE: Find $\int_{|z|=5} \frac{zf'(z)}{f(z)}dz$ when $24\le |f'''(z)|\le 30$, $f(0)=f(1)=f(2)=3$
QUESTION [6 upvotes]: Let $f$ be an entire function such that $24\le |f'''(z)|\le 30$, $f(0)=f(1)=f(2)=3$. I want to find
$$\int_{|z|=5} \frac{zf'(z)}{f(z)}dz.$$
My attempt: Since $f$ is entire, $f'''$ is also entire. $f'''$ is also bounded. Thus, by Liouville's theorem, $f'''$ is constant and $f$ is a cubic function. Since $f(0)=f(1)=f(2)=3$, $f(z)=az(z-1)(z-2)+3$ and $4\le|a|\le 5$.
To use the residue theorem, we should know the zeros of $f(z)=0$, but I don't know how to find them. Where is the way I have to go?

REPLY [4 votes]: Let $f$ be a nice function, and $C$ a nice contour (simple, positively oriented
at least). Consider
$$I=\int_C \frac{zf'(z)}{f(z)}\,dz.$$
As $f$ is nice, $f$ won't have any repeated zeros inside $C$. Let $a$ be a zero inside $C$. Then the residue of the integrand is
$$\lim_{z\to a}(z-a)\frac{zf'(z)}{f(z)}
=af'(a)\lim_{z\to a}\frac{z-a}{f(z)}
=a.$$
So $I$ is $2\pi i$ times the sum of the zeros of $f$ within the contour.
In your example, can you prove that the zeros of your cubic are within
your circle? If so then the sum of them is $3$ (Vieta) etc.