Is it possible to write each element of the unitary disk $\mathbb{D}={\{z\in\mathbb{C}: |z|<1}\}$ as $z=\text{tanh}(\frac{t}{2})e^{-i\theta}$?

Question

Is it possible to write each element of the unitary disk $\mathbb{D}={\{z\in\mathbb{C}: |z|<1}\}$ as $z=\tanh(\frac{t}{2})e^{-i\theta}$ with $t>0$?

It seems simple, but I don't see how to verify it. Thanks in advance for the help.

Answer 1

Let $z \in \Bbb{D}$. Then, in the polar representation, $z = r \mathrm{e}^{\mathrm{i} \varphi}$, where $|z| = r$ and $\varphi = \arg(z)$ satisfy $0 \leq r < 1$. (Typically, one also restricts $\varphi$ to some half-open interval with width $2\pi$, but we don't need to do so.)

The hyperbolic tangent function has $\tanh(0) = 0$, $\tanh'(x) > 0$, and $\lim_{x \rightarrow \infty} \tanh(x) = 1$. Consequently, for each $r \in [0,1)$, there is a unique $t/2$ such that $\tanh(t/2) = r$. Also, if we set $\theta = -\varphi$, we have $z = \tanh(t/2)\mathrm{e}^{-\mathrm{i}\theta}$ with $t \geq 0$.

If we restrict $t > 0$, then there is no way to represent the point $0 \in \Bbb{D}$. So on this point, the answer to your question is "no."

Answer 2

Hint:

$\tanh$ is an odd bijection from $\mathbf R$ onto $(-1,1)$. So you have to solve $\tanh\frac t2=|z|$.

It will be useful to remember that in inverse function $\operatorname{argtanh}$ can be expressed as a log.