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Put $p_n(z)=\sum_{k=0}^n\frac{z^k}{k!}$. Show that for any $r>0$ and any $n\ge 0$, there exists a point $z_0$ with $|z_0|=r$ such that $|p_n(z_0)|=|e^{z_0}|$.

This is actually the second part of a problem, and the first part used Rouche's theorem and was quite easy. I've gotten stuck, but I'm relatively sure I'll be using Rouche's again.

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up vote 2 down vote accepted

Let $r>0$ and $N \geq 1$. Let's suppose $|p_N(z)|=|e^z- \sum_{k=N}^{\infty} \frac{z^k}{k!}|<|e^z|$ holds for every $z$ with $|z|=r$. Rouche's Theorem says in this case that $e^z$ and $\sum_{k=N}^{\infty} \frac{z^k}{k!}$ have the same amount of zeros inside $D(0,r)$; since $|e^z|>0$ for every $z$ and $\sum_{k=N}^{\infty} \frac{0^k}{k!}=0$ this is a contradiction. So there must be a $z_1$ with $|z_1|=r$ such that $|e^{z_1}- \sum_{k=N}^{\infty} \frac{{z_1}^k}{k!}|\geq|e^{z_1}|$. We also know $|p_N(z)|<|e^z|$ for every $z$ with Im$z$=0.

So by continuity on $\partial D(0,r)$ there is a point $z_0 \in \partial D(0,r)$ such that $$|p_N(z_0)|=|e^{z_0}|$$

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Thanks for your help! – Bey Jan 12 '13 at 0:39

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