User:Shahpour

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Theorem

Let $f$ and $g$ be complex-valued functions which are holomorphic in the interior of some simply connected region $D$.

Let $\left\vert{g \left({z}\right)}\right\vert < \left\vert{f \left({z}\right)}\right\vert$ on the boundary of $D$.


Then $f$ and $f + g$ have the same number of zeroes in the interior of $D$ counted up to multiplicity.


Proof

Let $N_f$ and $N_{f + g}$ be the number of zeroes of $f$ and $f + g$ in $D$ respectively.

By the Argument Principle:

$\displaystyle N_f = \frac 1 {2 \pi i} \oint_D \frac {f' \left({z}\right)} {f \left({z}\right)} \rd z$

Similarly:

$\displaystyle N_{f + g} = \frac 1 {2 \pi i} \oint_D \frac {\left({f + g}\right)' \left({z}\right)} {\left({f + g}\right) \left({z}\right)} \rd z$

We aim to show that $N_f = N_{f + g}$.

From $\left\vert{g \left({z}\right)}\right\vert < \left\vert{f \left({z}\right)}\right\vert$ we have that $f$ is non-zero on $D$, as $\left\vert{z}\right\vert \ge 0$ for all $z \in \C$.

From the fact that $\left\vert{g \left({z}\right)}\right\vert \ne \left\vert{f \left({z}\right)}\right\vert$ we also have that $g \left({z}\right) \ne - f \left({z}\right)$, so $f + g$ is also non-zero on $D$.

We have:

\(\displaystyle N_{f + g} - N_f\) \(=\) \(\displaystyle \frac 1 {2 \pi i} \oint_D \frac {\left({f + g}\right)' \left({z}\right)} {\left({f + g}\right) \left({z}\right)} \rd z - \frac 1 {2 \pi i} \oint_D \frac {f' \left({z}\right)} {f \left({z}\right)} \rd z\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \pi i} \oint_D \left({\frac {\left({f + g}\right)' \left({z}\right)} {\left({f + g}\right) \left({z}\right)} - \frac {f' \left({z}\right)} {f \left({z}\right)} }\right) \rd z\) Linear Combination of Contour Integrals
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \pi i} \oint_D \left({\frac {\left({f \left({1 + \frac g f}\right)}\right)' \left({z}\right)} {\left({f \left({1 + \frac g f}\right)}\right) \left({z}\right)} - \frac{f' \left({z}\right)} {f \left({z}\right)} }\right) \rd z\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \pi i} \oint_D \left({\frac {\left({f' \left({1 + \frac g f}\right)}\right) \left({z}\right)} {\left({f \left({1 + \frac g f}\right)}\right) \left({z}\right)} + \frac {\left({f \left({1 + \frac g f}\right)'}\right)\left({z}\right)} {\left({f \left({1 + \frac g f}\right)}\right) \left({z}\right)} - \frac {f' \left({z}\right)} {f \left({z}\right)} }\right) \mathrm dz\) Product Rule

As $g \left({z}\right) \ne - f \left({z}\right)$, we have $\dfrac {g \left({z}\right)} {f \left({z}\right)} \ne -1$, so $1 + \dfrac g f$ has no zeroes on $D$.

So:

\(\displaystyle \frac 1 {2 \pi i} \oint_D \left({\frac {\left({f' \left({1 + \frac g f}\right)}\right) \left({z}\right)} {\left({f \left({1 + \frac g f}\right)}\right) \left({z}\right)} + \frac {\left({f \left({1 + \frac g f}\right)'}\right)\left({z}\right)} {\left({f \left({1 + \frac g f}\right)}\right) \left({z}\right)} - \frac {f' \left({z}\right)} {f \left({z}\right)} }\right) \mathrm dz\) \(=\) \(\displaystyle \frac 1 {2 \pi i} \oint_D \left({\frac {f' \left({z}\right)} {f \left({z}\right)} + \frac {\left({1 + \frac g f}\right)' \left({z}\right)} {\left({1 + \frac g f}\right) \left({z}\right)} - \frac {f' \left({z}\right)} {f \left({z}\right)} }\right) \rd z\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \pi i} \oint_D \frac {\left({1 + \frac g f}\right)' \left({z}\right)} {\left({1 + \frac g f}\right) \left({z}\right)} \rd z\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)

Hence $N_{f + g} = N_f$.

$\blacksquare$


Source of Name

This entry was named for Eugène Rouché.

Theorem

Rouché's Theorem for analytic functions.

If $f(z)$ and $g(z)$ 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