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KoenigsFunction

Stephen Crowley edited this page Mar 2, 2023 · 22 revisions

To prove that $f(z)$ is the Koenigs function of the hyperbolic tangent, we need to show that it satisfies the following properties:

$f(z)$ is entire.

$f(z)$ maps the upper half-plane to the unit disk.

$f(z)$ has a simple pole at $\frac{\pi i}{2}$ and residue $1$.

$f(z)$ has the functional equation $f(z+\pi i) = -f(z)$.

Let's start by showing that $f(z)$ is entire. We can see that $f(z)$ is entire because it is a product of two entire functions: $\sinh(\pi z)$ and $\sinh(\pi/2)$. This satisfies the first property.

Next, we will show that $f(z)$ maps the upper half-plane to the unit disk. First, note that $\sinh(\pi z)$ maps the upper half-plane to the entire complex plane, and $\sinh(\pi/2)$ maps the upper half-plane to the positive real axis. Therefore, $f(z)$ maps the upper half-plane to the complex plane without the negative real axis. We can then apply a Möbius transformation $T(z) = \frac{z-i}{z+i}$, which maps the upper half-plane to the unit disk, to obtain $T(f(z)) = \frac{\sinh(\pi z) - i\sinh(\pi/2)}{\sinh(\pi z) + i\sinh(\pi/2)}$. We can check that this maps the upper half-plane to the unit disk using the fact that $|\sinh(\pi z)| > |\sinh(\pi/2)|$ for all $z$ in the upper half-plane.

Next, we will show that $f(z)$ has a simple pole at $\frac{\pi i}{2}$ with residue $1$. We can see that $\sinh(\pi z)$ has a simple zero at $\frac{\pi i}{2}$, and $\sinh(\pi/2)$ is nonzero, so $f(z)$ has a simple pole at $\frac{\pi i}{2}$ with residue $\frac{\sinh(\pi/2)}{\sinh(\pi/2)} = 1$.

Finally, we will show that $f(z+\pi i) = -f(z)$. We can see that $\sinh(\pi z)$ has period $i\pi$, and $\sinh(\pi/2)$ is a constant, so $f(z+\pi i) = \frac{\sinh(\pi(z+\pi i))}{\sinh(\pi/2)} = \frac{\sinh(\pi z)\cosh(\pi i) + \cosh(\pi z)\sinh(\pi i)}{\sinh(\pi/2)} = -\frac{\sinh(\pi z)}{\sinh(\pi/2)} = -f(z)$. This satisfies the fourth property.

Therefore, we have shown that $f(z)$ satisfies all four properties of a Koenigs function of the hyperbolic tangent, and thus $f(z)$ is the Koenigs function of the hyperbolic tangent.

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