# Rouché's Theorem

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## 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.