Biholomorphic mapping of $\tan(z)$

I'm supposed to solve this question:

6. Show that the function $\tan$ maps the vertical strip $-\frac{\pi}{4}<x<\frac{\pi}{4}$ biholomorphically to $\dot B(0,1)$

It is obvious that $\tan(z)$ will be holomorphic in this vertical strip. To prove injectivity and surjectivity I tried to write: $$\tan(z) = \frac{\sin(2x)+\sinh(2y)i}{\cosh(2y)+\cos(2x)}=w$$ With $z=x+iy$ and $w=u+iv$, working this out I get: $$(\cosh(2y)+\cos(2x))u=\sin(2x)$$ $$(\cosh(2y)+\cos(2x))v=\sinh(2y)$$ I should get now that for $w$ in the open unit disc, there is a unique $z$ in the vertical strip, but I have no clue how to proceed. I hope someone can help me out. Please tell also if there is a better/faster way to prove bijectivity than my attempt.

Solution 1:

$$ \tan(z) = \frac {\sin z}{\cos z} = \frac {\frac {1}{2i}(e^{iz} - e^{-iz})}{\frac {1}{2}(e^{iz} + e^{-iz})} = \frac 1i \frac {e^{2iz} - 1}{e^{2iz} + 1} $$ is the composition of the thee mappings $$ z\to u = 2iz, \quad u \to v = e^u, \quad v \to w = \frac 1i\frac{u-1}{u+1} \, . $$

• The first maps the vertical strip $-\frac{\pi}{4}<x<\frac{\pi}{4}$ biholomorphically to the horizontal strip $-\frac{\pi}{2}<y<\frac{\pi}{2}$.
• The second maps the horizontal strip biholomorphically to the right half-plane.
• The third (a Möbius transformation) maps the right half-plane biholomorphically to the unit disk.