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Use Rouche's theorem to show that $e^z$ cannot vanish on the unit disk.

Generally, in applications of Rouche's theorem, if I was trying to prove $f$ had a certain number of zeros in the unit disk, I would find holomorphic functions $g$ and $h$ such that $g + h = f$, show that $|g| > |h|$ on the unit circle, and then conclude that $f$ had the same number of 0's as $g$. However, I 'm not sure what functions $g$ and $h$ I would use, in this case... I know that $e^z = e^x \cos y + ie^x \sin y$, but I can't show that either one of those terms is bounded by the other on the unit circle. Am I missing something, or should I be applying Rouche's theorem in a different way?

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    $\begingroup$ What a bizarre task (given that it's trivial from the formula for $e^z$ to see that it can't vanish anywhere)...! $\endgroup$ Jun 12, 2015 at 6:08
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    $\begingroup$ @HansLundmark : Which formula for $e^z$ do you have in mind? I can't say it seems trivial that $z\mapsto\sum_{n=1}^\infty z^n/n!$ never vanishes. ${}\qquad{}$ $\endgroup$ Jun 12, 2015 at 15:48
  • $\begingroup$ $1=e^z\cdot e^{-z}$ shows that $e^z$ doesn't vanish, I guess is what @HansLundmark was thinking. $\endgroup$ Jun 12, 2015 at 16:17
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    $\begingroup$ @MichaelHardy: I was talking about the formula quoted in the question, of course: $e^{x+iy}=e^x (\cos y + i \sin y)$. We know from real analysis that $e^x > 0$ and that $\cos y$ and $\sin y$ can't vanish simultaneously. $\endgroup$ Jun 12, 2015 at 19:32
  • $\begingroup$ @HansLundmark : ok, I'm convinced. $\endgroup$ Jun 12, 2015 at 21:02

2 Answers 2

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Take $f(z)=e^z$ and $g(z)=k$ with $k\in \Bbb{R},\,k\gt {e^2\over 2}$ we have $|e^z-k|\lt k$ on the unit circle and the exponential has the same number of zeros as the positive constant on the unit disk

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  • $\begingroup$ @copper.hat 3 Last time I checked $i\pi$ was not on the unit circle so your counterexample may work on the circle of radius $\pi$ but certainly not the unit circle $\endgroup$
    – marwalix
    Jun 12, 2015 at 19:13
  • $\begingroup$ Oops, excuse me, was confusing $e^{i \theta}$ with $e^{e^{i \theta}}$. $\endgroup$
    – copper.hat
    Jun 12, 2015 at 19:28
  • $\begingroup$ +1: Very succinct. Why did you choose ${e^2\over 2}$? (I think ${ e\over 2}$ will do.) $\endgroup$
    – copper.hat
    Jun 12, 2015 at 20:22
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    $\begingroup$ I solved the inequality $|e^{\cos\alpha}e^{i\sin\alpha}-k|\lt k$ that lead to $k\gt e^{2\cos\alpha}/2\cos(\sin\alpha)$ and the RHS reaches a maximum at $\alpha=2n\pi$ $\endgroup$
    – marwalix
    Jun 12, 2015 at 20:51
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Here is an answer if you are willing to accept the inequality $e-(1+1+{1 \over 2!})< {1 \over e}$.

If $|z|=1$, then $|e^z| = e^{\operatorname{re} z} \ge {1 \over e}$.

Let $p(z) = 1+z+{1 \over 2!} z^2$. If $|z| = 1$, then $|e^z-p(z)| = |{1 \over 3!} z^3+\cdots| \le {1 \over 3!}+\cdots = e-(1+1+{1 \over 2!})< {1 \over e} \le |e^z|$.

Hence $\exp$ and $p$ have the same number of zeros inside the unit circle.

It is straightforward to verify that the zeros of $p$ are $-1 \pm i$.

Addendum: Here is a proof of the initial inequality: First, note that $e-2 = {1 \over 2!}+\cdots < {1 \over 2} (1+{1 \over 2} + {1 \over 2^2} + \cdots) = 1$, and so $e < 3$. Then $e-2-{1 \over 2!} = {1 \over 3!}+\cdots < {1 \over 3!} (1+{1 \over 3} + {1 \over 3^2} + \cdots) = {1 \over 4}< {1 \over 3} < {1 \over e}$.

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