A bijective map between a horizontal strip and the unit disc.

Let $z=x+i \in \mathbb{R}+i$ with $x\in\mathbb{R}$. Then $$ \phi(x+i) = \tanh\left(\frac{\pi}{4}(x+i)\right). $$ Set $\frac{\pi}{4}x = u$, so that $u\in\mathbb{R}$. Then $$ \phi(x+i) = \tanh(u + i\tfrac{\pi}{4}). $$ Using the formula for the hyperbolic tangent of a complex argument, $$ \tanh(a+ib) = \frac{\tanh(a) + i\tan(b)}{1 + i\tanh(a)\tan(b)}, $$ we plug in $a=u$ and $b=\frac{\pi}{4}$, noting that $\tan(\frac{\pi}{4})=1$: $$ \phi(x+i) = \frac{\tanh(u) + i}{1 + i\tanh(u)}=\frac{\tanh(u) + i}{1 + i\,\tanh(u)} = \frac{2\,\tanh(u)}{1 + \tanh(u)^2} + i\,\frac{1 - \tanh(u)^2}{1 + \tanh(u)^2}. $$

We can easily check that 
\[(\frac{2\,\tanh(u)}{1 + \tanh(u)^2})^2+(\frac{1 - \tanh(u)^2}{1 + \tanh(u)^2})^2=1.\]
Hence $\phi$ maps $\mathbb{R}+i$ to the unit circle. 

Note that $\tanh (-\infty)=-1$ and $\tanh (\infty)=1$. Note also that the function
$$\frac{2\,\tanh(u)}{1 + \tanh(u)^2}$$
is a strcitly increasing function whose range is $(-1,1)$ with 
$$\phi(-\infty+i)=-1, \ \ \ \phi (\infty+i)=1,$$
and as $x$ goes from $-\infty$ to $+\infty$ the values of $\phi(-\infty+i)$ goes from $-1$ to $+1$ on the unit circle. Hence $\phi$ maps $\mathbb{R}+i$ bijectively to $$\{e^{i\theta} : \theta\in(0,\pi)\}.$$

Indeed, as $x$ runs from $-\infty$ to $\infty$, $\phi(x+i)$ runs along the unit circle from $-1$ to $1$ through $i$, which is the upper semicircle of the unit circle $\{e^{i\theta}:\theta\in(0,\pi)\}$. A similar argument, or using symmetry, shows that $\mathbb{R}-i$ maps to the lower semicircle $\{e^{i\theta}:\theta\in(-\pi,0)\}$.