# User talk:Shahpour

## Theorem

Rouché's Theorem for analytic functions.

If $f(z)$ and $g(z)$ be are analytic (holomorphic) functions inside and on a simple closed curve $C$ and if $|g(z)|<|f(z)|$ for $z\in C$, then $f(z)+g(z)$ and $f(z)$ have the same number of zeros inside $C$.

## Proof

Because $|f(z)|>|g(z)|\ge0$ on $C$ it follows that $|f(z)|\neq0$ and $|f(z)+g(z)|\neq0$ also. Let $N_1$ and $N_2$ be number of zeros of $f(z)$ and $f(z)+g(z)$, respectively, inside $C$. By argument's principle $\displaystyle N_1=\frac{1}{2\pi}\Delta_Carg[f(z)]$ and $\displaystyle N_2=\frac{1}{2\pi}\Delta_Carg[f(z)+g(z)]$ so \begin{eqnarray*}

 N_2 &=& \frac{1}{2\pi}\Delta_Carg[f(z)+g(z)] \\
&=& \frac{1}{2\pi}\Delta_Carg[f(z)][1+\frac{g}{f}(z)] \\
&=& \frac{1}{2\pi}\Delta_Carg[f(z)]+\frac{1}{2\pi}\Delta_Carg[1+\frac{g}{f}(z)]\\
&=& N_1+\frac{1}{2\pi}\Delta_Carg[1+\frac{g}{f}(z)]


\end{eqnarray*} Let $\displaystyle\omega=1+\frac{g}{f}(z)$ is a point in range of $\displaystyle1+\frac{g}{f}(z)$ that is on it's graph. From assumption $|g(z)|<|f(z)|$ we have $$|\omega-1|=\Big|\frac{g}{f}(z)\Big|<1$$ so $\omega$ must be inside the circle $|\omega-1|<1$ for $z\in C$, that shows $\omega$ doesn't meet $0$ then $\displaystyle\Delta_Carg[w]=\Delta_Carg[1+\frac{g}{f}(z)]=0$ and we conclude $N_2=N_1$.

## Theorem

Rouché's Theorem for harmonic maps. As Rouché's Theorem is proved for analytic functions, this theorem remains true for harmonic maps also. This theorem says:

If $f(z)$ and $f(z)+g(z)$ be sense-preserving harmonic maps in simple connected domain $D$, continuous in $\bar{D}$ and $|g(z)|<|f(z)|$ for $z\in\partial C$, then $f(z)+g(z)$ and $f(z)$ have the same number of zeros in $D$.

## Proof

From $|f(z)|>|g(z)|\ge0$ on $C$ it follows that $|f(z)|\neq0$ and $|f(z)+g(z)|\neq0$ also. Let $N_1$ and $N_2$ be number of zeros of $f(z)$ and $f(z)+g(z)$, respectively, in $D$. Since $f(z)$ and $f(z)+g(z)$ be sense-preserving harmonic maps by argument's principle for harmonic maps $\displaystyle N_1=\frac{1}{2\pi}\Delta_Carg[f(z)]$ and $\displaystyle N_2=\frac{1}{2\pi}\Delta_Carg[f(z)+g(z)]$ so \begin{eqnarray*}

 N_2 &=& \frac{1}{2\pi}\Delta_Carg[f(z)+g(z)] \\
&=& \frac{1}{2\pi}\Delta_Carg[f(z)][1+\frac{g}{f}(z)] \\
&=& \frac{1}{2\pi}\Delta_Carg[f(z)]+\frac{1}{2\pi}\Delta_Carg[1+\frac{g}{f}(z)]\\
&=& N_1+\frac{1}{2\pi}\Delta_Carg[1+\frac{g}{f}(z)]


\end{eqnarray*} Let $\displaystyle\omega=1+\frac{g}{f}(z)$ is a point in range of $\displaystyle1+\frac{g}{f}(z)$ that is on it's graph. From assumption $|g(z)|<|f(z)|$ we have $$|\omega-1|=\Big|\frac{g}{f}(z)\Big|<1$$ so $\omega$ must be inside the circle $|\omega-1|<1$ for $z\in C$, that shows $\omega$ doesn't meet $0$ then $\displaystyle\Delta_Carg[w]=\Delta_Carg[1+\frac{g}{f}(z)]=0$ and we conclude $N_2=N_1$.