= Solution
At a regular point, $f(z_0+\zeta)-f(z_0)=f'(z_0)\zeta+O(\zeta^2)$. Multiplication by a nonzero <complex number> rotates every tangent by the same angle and multiplies every <length> by the same positive factor. Oriented angles are preserved. If the first <derivative> vanishes but the second does not, the leading term is $f''(z_0)\zeta^2/2$: tangent arguments are doubled, so angles are doubled, modulo the identification of rays. This is a local statement; <holomorphy> alone does not imply global <injectivity>.
Write $z=x+iy$. Then
$$
w=\sin\frac{\pi x}{2}\cosh\frac{\pi y}{2}
+i\cos\frac{\pi x}{2}\sinh\frac{\pi y}{2}.
$$
For $-1<x<1$ and $y>0$, the imaginary part is positive. The bottom edge maps monotonically to $[-1,1]$; the left and right vertical edges map to $(-\infty,-1]$ and $[1,\infty)$ respectively. The inverse below establishes that every point of the open upper half-plane occurs exactly once. At fixed $y=a$, putting $u=\operatorname{Re}w$, $v=\operatorname{Im}w$, one obtains
$$
\boxed{\frac{u^2}{\cosh^2(\pi a/2)}+
\frac{v^2}{\sinh^2(\pi a/2)}=1,\qquad v\ge0.}
$$
This is the upper half of an <ellipse>, traversed from its left endpoint to its right endpoint.
Choose the branch $s(w)=\sqrt{1-w^2}$ analytic in the upper half-plane and positive on the boundary interval $(-1,1)$. Then the inverse branch into the strip is
$$
\boxed{z=-\frac{2i}{\pi}\log\bigl(iw+s(w)\bigr).}
$$
The <logarithm> is chosen continuously so that $-1<\operatorname{Re}z<1$ and $\operatorname{Im}z>0$; for instance $w=iv$, $v>0$, gives $z=(2i/\pi)\operatorname{arsinh}v$. Differentiation gives $dz/dw=2/(\pi\sqrt{1-w^2})$. The only zeros of the original <derivative> in the closed strip are $z=\pm1$, with images $w=\pm1$. The inverse has square-root <branch points> there, and cannot extend holomorphically across them. At either corner the original map doubles the right angle to a straight angle; a holomorphic inverse with nonzero <derivative> could not halve it. The open strip interior has no such failure.
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