# proof of Rouché’s theorem

Consider the integral

$$N(\lambda )=\frac{1}{2\pi i}{\oint}_{C}\frac{{f}^{\prime}(z)+\lambda {g}^{\prime}(z)}{f(z)+\lambda g(z)}\mathit{d}z$$ |

where $0\le \lambda \le 1$. By the hypotheses, the function $f+\lambda g$ is non-singular on $C$ or on the interior of $C$ and has no zeros on $C$. Hence, by the argument principle, $N(\lambda )$ equals the number of zeros (counted with multiplicity) of $f+\lambda g$ contained inside $C$. Note that this means that $N(\lambda )$ must be an integer.

Since $C$ is compact^{}, both $|f|$ and $|g|$ attain minima and maxima on $C$. Hence there exist positive real constants $a$ and $b$ such that

$$|f(z)|>a>b>|g(z)|$$ |

for all $z$ on $C$. By the triangle inequality^{}, this implies that $|f(z)+\lambda g(z)|>a-b$ on $C$. Hence $1/(f+\lambda g)$ is a continuous function^{} of $\lambda $ when $0\le \lambda \le 1$ and $z\in C$. Therefore, the integrand is a continuous function of $C$ and $\lambda $. Since $C$ is compact, it follows that $N(\lambda )$ is a continuous function of $\lambda $.

Now there is only one way for a continuous function of a real variable to assume only integer values – that function must be constant. In particular, this means that the number of zeros of $f+\lambda g$ inside $C$ is the same for all $\lambda $. Taking the extreme cases $\lambda =0$ and $\lambda =1$, this means that $f$ and $f+g$ have the same number of zeros inside $C$.

Title | proof of Rouché’s theorem^{} |
---|---|

Canonical name | ProofOfRouchesTheorem |

Date of creation | 2013-03-22 14:34:26 |

Last modified on | 2013-03-22 14:34:26 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 6 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 30E20 |