Zeros inside an annulus

julien_santini

Let $f$ be analytic in the annulus $A= \{z;\ 1 \leq |z| \leq 2 \}$ such that $|f|=5$ on the frontier of $A$ . Show that $f$ has at least $2$ zeroes in $A$ .

**Posted:** Tue Jan 03, 2006 8:34 pm

jmerry

It's boundary, not frontier.

A silly counterexample: $f(x)=5$ .
I suppose the problem really says "nonconstant". I've got it up to details, but the topology is really annoying.

It's not trivial to show such functions exist.

**Posted:** Tue Jan 03, 2006 10:39 pm

julien_santini

Yeah nonconstant. Well I found that such a function has at least one zero (this is rather trivial via the maximum modulus principle), but I still cannot see (even intuitively) where the second zero comes from.

**Posted:** Wed Jan 04, 2006 9:35 am

jmerry

Also, that should be "counting multiplicity"- I can construct an example that has one zero of multiplicity two.

Here's the nice argument:
Let $C_1$ be the inner circle and $C_2$ be the outer circle. By the argument principle, the number of zeros of $f$ (counting multiplicity) is $\int_{C_2}\frac{f'(z)}{2\pi i f(z)}\,dz-\int_{C_1}\frac{f'(z)}{2\pi i f(z)}\,dz$ . Each of those integrals must be an integer. Since $f$ is not constant, $\frac{f'}{f}$ is not uniformly zero on either curve.

Now, by geometric considerations, $z\frac{f'(z)}{f(z)}$ is pure real on both circles. In addition, since $|f(z)|<5$ inside the annulus by the maximum modulus principle, $z\frac{f'(z)}{f(z)}$ has positive (nonnegative, really) real part on $C_2$ and negative real part on $C_1$ (look at what happens to $f$ along the curve $tz$ ). This makes $\int_{C_2}\frac{f'(z)}{2\pi i f(z)}\,dz$ a positive integer and $\int_{C_1}\frac{f'(z)}{2\pi i f(z)}\,dz$ a negative integer, so the difference is at least 2.

**Posted:** Wed Jan 04, 2006 11:49 am

kyryk · New Member Offline Joined: 01 Apr 2006 Posts: 4

I stumbled upon this problem recently and couldn't figure it out. I am happy to see someone solved it Smile

Could you elaborate on it. Why is zf'/f real? Why is it important that the f'/f is not uniformly zero; even if it is not zero, then still the integrals could be. Any help would be appreciated!

**Posted:** Sat Apr 01, 2006 9:55 pm

jmerry

The argument isn't just that $z\frac{f'(z)}{f(z)}$ is real- it's that $z\frac{f'(z)}{f(z)}\ge 0$ on the outer circle and $z\frac{f'(z)}{f(z)}\le0$ on the inner circle. With that, the only way either integral can be zero is if $z\frac{f'(z)}{f(z)}$ is identically zero on the circle.

Why is $z\frac{f'(z)}{f(z)}$ purely real? Differentiate $\ln(f(re^{i\theta}))$ , where $r$ is the radius of one of the circles. The derivative is $ire^{i\theta}\frac{f'(re^{i\theta})}{f(re^{i\theta})}$ ; this is purely imaginary since the real part of $\ln(f(e^{i\theta}))$ is identically $\ln 5$ . Divide by $i$ and set $z=re^{i\theta}$ : $z\frac{f'(z)}{f(z)}$ is real when $z$ is on one of the circles.

There are other, more geometric ways of seeing this- look at the definition of the derivative along certain curves. To get the sign result, we use a radial line instead of the circle.

**Posted:** Sat Apr 01, 2006 10:31 pm

kyryk · New Member Offline Joined: 01 Apr 2006 Posts: 4

Hi,
Thank you for your reply! I have a few more questions. How do you change your argument if f(z) is a negative number? Do you change the branch cut for the log function? Also, how do you know exactly that zf'/f is positive/negative? I am not able to see this. Also, how do you go from zf'/f to dz f'/f? Do you just use the fact that z~i dz?
Thank you very much, and sorry to bother you with all the details!

**Posted:** Sat Apr 01, 2006 10:47 pm

kyryk · New Member Offline Joined: 01 Apr 2006 Posts: 4

Actually, I think I got it! Thank you for your help!

In one of your previous posts you mention that "the topology is really annoying" Is there a general approach to this kind of questions? Could you recommend a book?

Also, can you give a function with a double root? How do you construct one?

Once again, thank you for your help. THis question really intrigued me and that's why I have so many questions.

**Posted:** Sat Apr 01, 2006 11:04 pm