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 $N_1=\dfrac{1}{2\pi}\Delta_Carg[f(z)]$ and $N_2=\dfrac{1}{2\pi}\Delta_Carg[f(z)+g(z)]$ 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 $\omega=1+\dfrac{g}{f}(z)$ is a point in range of $1+\dfrac{g}{f}(z)$ 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 $\Delta_Carg[w]=\Delta_Carg[1+\dfrac{g}{f}(z)]=0$ and we conclude $N_2=N_1$.
Also see
http://en.wikipedia.org/wiki/Rouch%C3%A9's_theorem
http://mathworld.wolfram.com/RouchesTheorem.html
Sources
Rouché's Theorem for harmonic maps
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 $N_1=\dfrac{1}{2\pi}\Delta_Carg[f(z)]$ and $N_2=\dfrac{1}{2\pi}\Delta_Carg[f(z)+g(z)]$ 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 $\omega=1+\dfrac{g}{f}(z)$ is a point in range of $1+\dfrac{g}{f}(z)$ 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 $\Delta_Carg[w]=\Delta_Carg[1+\dfrac{g}{f}(z)]=0$ and we conclude $N_2=N_1$.