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Spring 2011 - Problem 12

calculus

Let $\Omega = \C \setminus \poc{-\infty, 0}$ and let $\log{z}$ be the branch of the complex logarithm on $\Omega$ that is real on the positive real axis (and analytic throughout $\Omega$ ). Show that for $0 < t < \infty$ , the number of solutions $z \in \Omega$ to

\log{z} = \frac{t}{z}

is finite and independent of $t$ .

Solution.

We will use polar coordinates: $z = re^{i\theta}$ with $\theta \in \p{-\pi, \pi}$ , which transforms the equation into

\log{r} + i\theta = \frac{t}{r}\p{\cos\theta - i\sin\theta}.

First, if we equate the imaginary parts, we get

\theta = -\frac{t\sin\theta}{r}.

Since $r, t > 0$ , we see that if $\theta \neq 0$ , then the left-hand side and right-hand side have different signs. Thus, the only possible roots occur with $\theta = 0$ . For the real parts, at $\theta = 0$ we have

\log{r} = \frac{t}{r} \iff f\p{r} \coloneqq \log{r} - \frac{t}{r} = 0.

Observe that

f'\p{t} = \frac{1}{r} + \frac{t}{r^2} = \frac{r + t}{r^2} > 0,

so $f$ is injective on $\p{0, \infty}$ . Since $f\p{r} \to \infty$ as $r \to \infty$ and if $0 < r < 1$ , then $f\p{r} < 0$ . Hence, by the mean value theorem, $f\p{r}$ has a root and by injectivity, this root is unique. Thus, the number of solutions to the equation is precisely $1$ for any $t > 0$ .