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I understand the statement of Rouché's theorem which (very basically) says that if we have a closed anticlockwise contour where functions $f$ and $g$ are holomorphic on a region $R$, and $f$ is greater than $g$, then $f+g$ will have the same number of zeroes as $f$.

However, I fail to see how to use it.

For example:

$f\left( z\right) :=z^{5}-6z+4$

By comparing it with

$g\left( z\right) =z^{5}$ on $\left\{ z\in \mathbb{C} :\left| z\right| =2\right\}$

Can someone show me how to apply Rouché's theorem here.

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Let $h(z)=-6z+4$. Then on $\{z\in\mathbb{C}: |z|=2\}$ you have $$|g(z)|=2^5=32>16=12+4\geq|-6z+4|=|h(z)|.$$ The functions $h$ and $g$ are holomorphic inside $\{z\in\mathbb{C}: |z|=2\}$, so by applying Rouché's Theorem you get that $h+g=f$ has the same number of zeros as $g$ inside the circle. Since $g(z)=z^5$ has obviously a zero of multiplicity $5$ at $0$, the function $f(z)$ has $5$ zeros in $\{z\in\mathbb{C}: |z|<2\}$.

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