Gamelin Exercise Problem: (Ch VIII.2):

Let $f$ & $g$ be analytic functions in an open set containing a circle $C$ & its interior. Suppose $|f(z) + g(z)| \lt |f(z)| + |g(z)| ; \forall z \in C$ . Show that $f$ & $g$ have the same number of zeros inside the circle $C$ counting multiplicity.

Clearly, it seemed to use Rouche's Theorem. But, there I faced trouble; since: I couldn't identify the "SMALL FUNCTION" & "BIG FUNCTION" as in the theorem of Rouche. So, please help!!


1 Answer 1


Let $V$ the open subset of $\mathbb{C}$ where $f,g$ are analytic; one can suppose that the circle has its center at $0$. I think that your hypothesis is $|f(z)-g(z)|<|f(z)|+|g(z)|$ for $z\in T=\{z; |z|=r\}$; also, we can replace $g$ by $-g$ wlog.

a) Let $U= \{z\in V; |f(z)-g(z)|<|f(z)|+|g(z)|\}$. Then $U$ is an open subset of $V$, containing $T$.

b) If $a\in U$ and for example $f(a)=0$, then $|g(a)|<|g(a)|$, a contradiction; the same for $g$. Hence $f(z)g(z)\not =0$ on $U$. The fonction $\displaystyle h(z)=\frac{f(z)}{g(z)}$ is hence analytic and non zero on $U$.

c) Let $\log $ the principal determination of the logarithm, defined on $W=\mathbb{C}\setminus{]-\infty,0]}$. If $h(z)=-t$, with $t\geq 0$, we get $1+t=|-t-1|<1+t$, a contradiction. Hence $h(z)\in W$ for all $z\in U$, and $F(z)=\log h(z)$ is analytic in $U$, and $\displaystyle F^{\prime}(z)=\frac{f^{\prime}(z)}{f(z)}-\frac{g^{\prime}(z)}{g(z)}$.

d) Now $$\frac{1}{2i\pi}\int_T F^{\prime}(z)dz=0=\frac{1}{2i\pi}\int_T \frac{f^{\prime}(z)}{f(z)}dz-\frac{1}{2i\pi}\int_T \frac{g^{\prime}(z)}{g(z)}dz$$ e) As $\displaystyle \frac{1}{2i\pi}\int_T \frac{f^{\prime}(z)}{f(z)}dz$ is the number of zeros of $f$ in the open disk $D(0,r)$, and $\displaystyle \frac{1}{2i\pi}\int_T \frac{g^{\prime}(z)}{g(z)}dz$ is the number of zeros of $g$ in the open disk $D(0,r)$ (counting multiplicities), we are done.

  • $\begingroup$ Yup!! .. Thanks for introducing the branch cut!! $\endgroup$
    – user86511
    Commented Sep 13, 2014 at 2:54

You must log in to answer this question.

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