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

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 $N_2=N_1$.