7
$\begingroup$

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?

$\endgroup$

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Very nice indeed. (+1) $\endgroup$
    – MSDG
    Aug 24, 2018 at 6:11
  • $\begingroup$ So for this problem we acn not find the exact value of the integral, because we can't find the constant $a$ in the OP's polynomial !! $\endgroup$
    – Empty
    Aug 25, 2018 at 7:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .