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Spent some time trying to tackle this problem. It is supposed to use Rouche's Theorem, but not sure how.

Show that $ze^z = a$ for $a \neq 0$ has infinitely many roots.

Rouches:
(1) $f$ and $g$ analytic in and on simple, closed $C$

(2) $|f|>|g|$ on $C$

(*) $f+g$ and $f$ have the same number of zeroes inside $C$

Michael Hardy
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J_Lopez8
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    It would be trivial using Picard's theorem. Does it have to be Rouché's theorem? – Daniel Fischer Jul 14 '14 at 20:43
  • I would like to see how to use Rouche's; I've seen something similar with Picard's. Either way, I'm interested, if you care to post. @DanielFischer – J_Lopez8 Jul 14 '14 at 21:04
  • So far, I'm drawing a blank on Rouché's theorem. Maybe I'll have an idea after turning off the computer. – Daniel Fischer Jul 14 '14 at 21:59
  • Tried something like this: $f = ze^z - z^2$ and $g = z^2 - a$, which yields $f+g = ze^z - a$. And knowing externally that $e^z =z$ has infinitely many roots, this could work, with one exception that we need to satisfy (2). I'm tabling this problem for now. – J_Lopez8 Jul 14 '14 at 22:39

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