User:Shahpour
Theorem
Let $\gamma$ be a closed contour.
Let $D$ be the region enclosed by $\gamma$.
Let $f$ and $g$ be complex-valued functions which are holomorphic in $D$.
Let $\cmod {\map g z} < \cmod {\map f z}$ on $\gamma$.
Then $f$ and $f + g$ have the same number of zeroes in $D$ 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:
- $\ds N_f = \frac 1 {2 \pi i} \oint_\gamma \frac {\map {f'} z} {\map f z} \rd z$
Similarly:
- $\ds N_{f + g} = \frac 1 {2 \pi i} \oint_\gamma \frac {\map {\paren {f + g}'} z} {\map {\paren {f + g} } z} \rd z$
We aim to show that $N_f = N_{f + g}$.
From $\cmod {\map g z} < \cmod {\map f z}$ we have that $f$ is non-zero on $\gamma$, otherwise we would have $\cmod {\map g z} < 0$.
From the fact that $\cmod {\map g z} \ne \cmod {\map f z}$ we also have that $\map g z \ne - \map f z$, so $f + g$ is also non-zero on $\gamma$.
We have:
\(\ds N_{f + g} - N_f\) | \(=\) | \(\ds \frac 1 {2 \pi i} \oint_\gamma \frac {\map {\paren {f + g}'} z} {\map {\paren {f + g} } z} \rd z - \frac 1 {2 \pi i} \oint_\gamma \frac {\map {f'} z} {\map f z} \rd z\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 \pi i} \oint_\gamma \paren {\frac {\map {\paren {f + g}'} z} {\map {\paren {f + g} } z} - \frac {\map {f'} z} {\map f z} } \rd z\) | Linear Combination of Contour Integrals | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 \pi i} \oint_\gamma \paren {\frac {\map {\paren {f \paren {1 + \frac g f} }'} z} {\map {\paren {\map f {1 + \frac g f} } } z} - \frac {\map {f'} z} {\map f z} } \rd z\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 \pi i} \oint_\gamma \paren {\frac {\map {\paren {\map {f'} {1 + \frac g f} } } z} {\map {\paren {\map f {1 + \frac g f} } } z} + \frac {\map {\paren {\map f {1 + \frac g f}'} } z} {\map {\paren {\map f {1 + \frac g f} } } z} - \frac {\map {f'} z} {\map f z} } \rd z\) | Product Rule for Derivatives |
So:
\(\ds \frac 1 {2 \pi i} \oint_\gamma \paren {\frac {\map {\paren {f' \paren {1 + \frac g f} } } z} {\map {\paren {f \paren {1 + \frac g f} } } z} + \frac {\map {\paren {f \paren {1 + \frac g f}'} } z} {\map {\paren {\map f {1 + \frac g f} } } z} - \frac {\map {f'} z} {\map f z} } \rd z\) | \(=\) | \(\ds \frac 1 {2 \pi i} \oint_\gamma \paren {\frac {\map {f'} z} {\map f z} + \frac {\map {\paren {1 + \frac g f}'} z} {\map {\paren {1 + \frac g f} } z} - \frac {\map {f'} z} {\map f z} } \rd z\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 \pi i} \oint_\gamma \frac {\map {\paren {1 + \frac g f}'} z} {\map {\paren {1 + \frac g f} } z} \rd z\) |
For brevity, write:
- $F = 1 + \dfrac g f$
As $\cmod {\dfrac g f} < 1$ on $\gamma$, we must have:
- $\cmod {\map \Re {\dfrac g f} } < 1$
That is:
- $0 < \map \Re F < 2$
on $\gamma$.
That is, the image of $\gamma$ under $F$ does not encircle $0$.
So, by the definition of winding number, we have:
- $\map {\mathrm {Ind}_{\map F \gamma} } 0 = 0$
So:
\(\ds \frac 1 {2 \pi i} \oint_\gamma \frac {\map {F'} z} {\map F z} \rd z\) | \(=\) | \(\ds \frac 1 {2 \pi i} \oint_{\map F \gamma} \frac 1 z \rd z\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\mathrm {Ind}_{\map F \gamma} } 0\) | Definition of Winding Number | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
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 $N_1=\dfrac{1}{2\pi}\Delta_Carg[f(z)]$ 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