2
$\begingroup$

I am practicing exam questions for my own exam in complex analysis. This was one I couldn't figure out.

Let $U = \mathbb{C} \setminus \{ x \in \mathbb{R} : x \leq 0 \} $ en let $\log : U \to \mathbb{C} $ be the usual holomorphic branch of the logarithm on $U$ with $\log (1) = 0$. Consider the function given by $f(z) = \log(z) - 4(z-2)^2$.

Q1: Show that $f$ has exactly two zeroes (including multiplicity) in the open disk $D(2,1) = \{ z \in \mathbb{C} : |z-2| < 1 \} $.

Q2: Show that $f$ has exactly two different zeroes in $D(2,1)$.

I strongly suspect we should use Rouché's Theorem for this. I tried to apply it by setting $h(z) = \log(z)$ and $g(z) = -4 (z-2)^2$. If $|z-2| = 1$, then $z = 2 + e^{it}$ with $t \in [0,2 \pi ) $. Then we have $|h(z)| = |\log(z)| \leq \log|z| = \log|2+e^{it}| \leq \log(|2| + |e^{it}|) = \log|2|\cdot \log|e^{it}| = \log(2) \cdot \log(1) = 0$.

Furthermore, we have $g(z) = |-4 (z-2)^2| = |-4| |z-2|^2 = 4$. So $|h(z)| < |g(z)|$ when $|z-2| = 1$. According to Rouché's Theorem, $f$ en $g$ have the same number of zeroes within $\{ |z-2| < 1 \}$, so we have to count the zeros of $g$ to find the number of zeroes of $f$. However, I can only find one zero of $g$, which is at $z=2$. Can you tell what's going wrong with my approach?

$\endgroup$
  • $\begingroup$ Don't forget to count the multiplicity of the zero of $g$ $\endgroup$ – user47693 Jun 17 '13 at 13:41
  • $\begingroup$ @LordBellington aha, the multiplicity of the zero of $g$ is two, so I guess I answered Q1 correctly after all. Do you know how to answer Q2? $\endgroup$ – Max Muller Jun 17 '13 at 13:46
4
$\begingroup$

Your estimate $|h(z)|\le 0$ (when $|z-2|=1$) cannot possibly be true: a nonconstant holomorphic function cannot be equal to zero on a circle. I marked the incorrect steps in red: $$|\log(z)| \color{red}{\leq} \log|z| = \log|2+e^{it}| \leq \log(|2| + |e^{it}|) \color{red}{=} \log|2|\cdot \log|e^{it}| = \log(2) \cdot \log(1) = 0$$

A correct estimate could look like $$|\log(z)| \le |\operatorname{Re} \log z| + |\operatorname{Im} \log z| = \log |z| + |\arg z| \le \log 3+ \pi/2 <3$$ which is not sharp but suffices for the application of Rouché's theorem, which settles Q1.

The question Q2 is less standard. One way to answer it is to observe that the real function $f(x)=\log x-4(x-2)^2$ has two distinct zeros on the interval $(1,3)$, because $f(1)<0$, $f(2)>0$, and $f(3)=\log 3-4<0$. Since we already know there are two zeros in $D(2,1)$ with multiplicity, there are no others.

$\endgroup$
  • $\begingroup$ Thanks! This is great. Someone mentioned one can answer Q2 by means of the deriviative of $f$, do you know how that might be useful? $\endgroup$ – Max Muller Jun 18 '13 at 14:09
  • 1
    $\begingroup$ @MaxMuller Yes, that is also an option. To solve Q2, we must rule out the possibility of a double zero. At a double zero of $f$, the derivative $f'(z)=z^{-1}-8(z-2)$ also vanishes. The equation $f'(z)=0$ is easy to solve with quadratic formula. One can find its only solution in $D(2,1)$ and check that $f$ does not turn to zero at that point. ... I considered this approach, but the one I gave in the answer seemed easier. $\endgroup$ – ˈjuː.zɚ79365 Jun 18 '13 at 14:12

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.