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Show that as the positive integer $N$ tends to $\infty$, the change in argument of $e^z − z$ is bounded on $3$ sides of the square with corners $ \pm 2\pi N$ $\pm 2\pi iN$ but is unbounded on the fourth side.

Show that $e^z = z$ has infinitely many complex roots.

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What's the contribution of the two exclamation marks to the title? –  joriki Jan 12 '13 at 8:21
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Maybe it's a double-factorial argument of some kind. –  AndrewG Jan 12 '13 at 8:24
    
(joking, of course) –  AndrewG Jan 12 '13 at 8:25
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looks like homework question –  Ram Jan 12 '13 at 8:38
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What have you tried? Which of the sides do you think will give the large change in argument? –  mrf Jan 12 '13 at 8:39
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1 Answer

  • On either horizontal side, $\arg z$ changes by $\pi/2$ while $\arg e^z$ is constant due to $e^{x+iy}=e^xe^{iy}$.

  • On the left side, $\operatorname{Re}(e^z-z) \ge -|e^z|+2\pi N=e^{-2\pi N}+2\pi N>0$.

  • On the right side, $e^z$ traverses the circle of radius $e^{2\pi N}$ exactly $2N$ times, and the term $z$, having the modest modulus $|z|\le N\sqrt{2}$, can't do much about that.

The conclusion about roots follows from the argument principle.

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