2

This problem came from Schaum's complex variable p.156

Suppose $f(x)$ and $(g)$ are analytic inside and on a simple closed curve $C$ and suppose $|g(z)|\lt |f(z)|$ on $C$. Then $f(z)+g(z)$ and $f(z)$ have the same number of zeros inside $C$.

Here is the proof:

Let $F(z)=\frac{g(z)}{f(z)}$, or briefly $g=fF$. then, if $N_1$ and $N_2$ are the number of zeros inside $C$ of $f+g$ and $f$, respectively, we have by Argument Theorem, using the fact that these function have no poles inside C, $$N_1=\frac1{2\pi i}∮\frac{f'+g'}{f+g}dz$$ $$N_2=\frac{1}{2\pi i}∮\frac{f'}fdz$$ then$$N_1-N_2=\frac 1{2\pi i}∮\frac{f'+f'F+fF'}{f+fF}dz-\frac 1{2\pi i}∮\frac{f'}{f}dz$$$$=\frac 1{2\pi i}∮\frac{f'(1+F)+fF'}{f(1+F)}dz-\frac 1{2\pi i}∮\frac{f'}{f}dz$$$$=\frac 1{2\pi i}∮\left\{\frac{f'}{f}+\frac{F'}{1+F}\right\}dz-\frac 1{2\pi i}∮\frac{f'}{f}dz=∮\frac{F'}{1+F}dz$$$$=\frac 1{2\pi i}∮F'(1-F+F^2-F^3+...)dz=0$$

using the given fact that $|F|\lt 1$ on $C$ so that the series is uniformly convergent on $C$ and term by term integration yields the value zero. Thus, $N_1=N_2$ as required.

I don't understand the last step that why it yields to zero. I also don't understand the wiki version.

Lord_Farin
  • 17,743
Y.H. Chan
  • 2,612
  • 1
    Hint: $F'F^n dz= \frac{1}{n+1}d F^{n+1}$ – Ma Ming Apr 25 '13 at 11:08
  • Sorry that I am not familiarize with complex analysis since I am a Secondary 5 student. So I don't know how to deal with the integral with $dF^{n+1}$ – Y.H. Chan Apr 25 '13 at 11:26
  • @Chan It is Exact, so the integral is zero. Alternatively, you can use http://en.wikipedia.org/wiki/Morera%27s_theorem – Ma Ming Apr 25 '13 at 11:31
  • $F(z)$ is analytic on $C$ but is it analytic inside $C$? Since although $f$ cannot be zero on $C$, it can be zero inside, which makes $F$ become a singularity point. If $F$ is not analytic, why the integral vanish to zero? – Y.H. Chan Apr 25 '13 at 13:26

0 Answers0