User talk:Shahpour

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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 $N_2=N_1$.

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 $\displaystyle N_1=\frac{1}{2\pi}\Delta_Carg[f(z)]$ 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 $N_2=N_1$.

Also see

Sources