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Homeproof of Rouch\'e's theorem

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# proof of Rouché’s theorem

Consider the integral

$N(\lambda)={1\over 2\pi i}\oint_{C}{f^{{\prime}}(z)+\lambda g^{{\prime}}(z)% \over f(z)+\lambda g(z)}dz$ |

where $0\leq\lambda\leq 1$. 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 $0\leq\lambda\leq 1$ 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 $C$.

## Mathematics Subject Classification

30E20*no label found*

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## Comments

## Unclear line in proof

The proof contains the line

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

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

for all z in C."

I don't follow this. Why should the maximum of |g| be less than the minimum of |f|? We only know that |f(z)|>|g(z)| at each z, and there's no reason I can see why the maximum of |g| should be at the same point as the minimum of |f|.