Let $f(z) = \sum_0^\infty{a_n z_0^n}$ be a complex power series with radius of convergence $R>0$ and let $z_0 \epsilon \, \mathcal{U}_R(0)$ an arbitrary point.

I need to show with $Rouché$ :
For every open neighbourhood $z_0 \epsilon \, \mathcal{U} \subset \mathcal{U}_R(0)$ there exists a $z_1 \epsilon \, \mathcal{U}$ and a natural $N$, so that $f(z_0) = \sum_0^\infty a_nz_0^n = \sum_0^N a_nz_1^n $
A hint is given by: Let $f$ be not constant. Then you can find an $\varepsilon >0$ and a $\delta >0$ so that $\mathcal{U}_\varepsilon(z_0) \subset \mathcal{U}$ and $|f(z_0)-f(z)| \geq \delta$ for all complex $z$ with $|z_0 - z| = \varepsilon $.

Thank you!


Let's subtract $f(z_0)$ from $f$, in other words consider the function $g(z)=f(z)-f(z_0)$ instead. This helps because Rouché's theorem is about zeros.

Since $g$ is not identically zero, it can be written as $(z-z_0)^k h(z)$ where $h(z_0)\ne 0$. Then $h\ne 0$ in a neighborhood of $z_0$, which is how you get the existence of $\epsilon,\delta$ mentioned in the hint.

Our goal is to show that the function $\tilde g(z) = \sum_{n=1}^N z^n$ has at least one zero in the $\epsilon$-neighborhood of $z_0$. So, we need $N$ such that $|g-\tilde g|<\delta$ in this neighborhood. And $g- \tilde g$ is just the tail of the Taylor series of $g$.


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