22 - Rouché's Theorem

Rouché's Theorem

Suppose $f$ and $g$ are holomorphic functions on an open set $U$ in $\mathbb{C}$
Let $\gamma$ be a simple contour that is contained inside of $U$ together with it’s interior

If $|f(z)|>|g(z)|$ for all $z\in {\gamma }^{\ast }$then

$$f\text{ and }f+g$$

have the same change in argument around $\gamma$

Hence they have the same number of $0$s (counted with multiplicity) inside of $\gamma$

Proof

Consider function

$$h=\frac{f+g}{f}=1+\frac{g}{f}$$

then the zeros of $h$ are zeros of $f+g$ and poles are zeros of $f$

Note that the assumption $|f(z)|>|g(z)|$ implies that $h$ has no zeroes or poles on ${\gamma }^{\ast }$

By the argument principle then
Difference between number of zeroes and poles is equal to winding number of

$$\Gamma (t)=h(\gamma (t))\text{ about zero}$$

By assumption for $z\in {\gamma }^{\ast }$ then $|g(z)|<|f(z)|$ so $\left|\frac{g(z)}{f(z)}\right|<1$

Hence image of $\Gamma$ lies entirely in $B(1,1)$ and thus lies in half plane $\{z:\mathrm{Re}(z)>0\}$
So picking the principal branch of $[\log ]$ defined on the half plane then integral

$${\int }_{\Gamma }\frac{1}{z}dz=\mathrm{Log}(h(\gamma (1)))-\mathrm{Log}(h(\gamma (0)))=0$$