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Fall 2018 - Problem 12

Rouché's theorem

If $\alpha \in \C$ satisfies $0 < \abs{\alpha} < 1$ and if $n \in \N = \set{1, 2, 3, \ldots}$ show that the equation

e^z\p{z - 1}^n = \alpha

has exactly $n$ simple roots int he half-plane $\set{z \mid \Re{z} > 0}$ .

Solution.

Consider the homotopy $f_t\p{z} = e^{tz} \p{z - 1}^n$ . Observe that on the circle $\abs{z - 1} = 1$ , we have

\abs{f_t\p{z}} = e^{t\Re{z}} \geq e^0 = 1 > \abs{\alpha},

so $f_t\p{z} - \alpha$ never vanishes on this circle. Thus, by Cauchy's theorem, if $\gamma$ is the curve $\abs{z - 1} = 1$ , we get

\frac{1}{2\pi i} \int_\gamma \frac{f_t'\p{z}}{f_t\p{z} - \alpha} = \frac{1}{2\pi i} \int_{f_t\p{\gamma}} \frac{\diff{w}}{w - \alpha}

is a constant. When $t = 0$ , we get $f_0\p{z} = \p{z - 1}^n$ , so the integral is equal to $n$ , and by the argument principle, this means that $e^z\p{z - 1}^n = \alpha$ has precisely $n$ solutions in the interior of the circle $\abs{z - 1} = 1$ .

To show that these solutions are simple, suppose $w$ is a solution to the equation. Then if $f\p{z} = e^z\p{z - 1}^n$ , then

f'\p{z} = e^z\p{z - 1}^n + ne^z\p{z - 1}^{n-1} \implies f'\p{w} = \alpha + \frac{n\alpha}{w - 1}.

This is $0$ if and only if

1 + \frac{n}{w - 1} = 0 \implies w = 1 - n,

but $\Re\p{1 - n} \leq 0$ , but $\Re{w} > 0$ . Thus, any solution to the equation must be simple.