Mapping of sin(z) Consider the strip $S=\{(x,y):-\pi/2<x<0,\ y>0\}$. Now, I've shown that
$$u+iv=\sin z=\sin x\cosh y+i\cos x\sinh y$$
Thus, because $-\pi/2<x<0$, $\sin x$ is negative and $\cos x$ is also negative. Because $y>0$, $\sinh y$ is positive and $\cosh y$ is also positive (greater than 1). Therefore, two facts:


*

*$\sin x\cosh y$ is negative.

*$\cos x\sinh y$ is negative.


These two facts tell me that all
$$u+iv=\sin z=\sin x\cosh y+i\cos x\sinh y$$
lie in the second quadrant. Now, here is my question. How can I formally prove that $\sin z$ maps $S$ one-to-one and onto the second quadrant?
Here is some Matlab code that indicates this is true.
[x,y]=meshgrid(linspace(-pi/2,0,41),linspace(0,5,41));
h=zeros(size(x));
subplot(1,2,1)
mesh(x,y,h), view(2)
z=complex(x,y);
f=@(z) sin(z);
w=f(z); u=real(w); v=imag(w);
subplot(1,2,2)
mesh(u,v,h), axis([-40,0,0,40]), view(2)


Thanks.
P.S.: I should also say that I have shown that most of the vertical lines are mapped onto branches of a hyperbola and most of the horizontal lines are mapped onto ellipses.
 A: Assume $\sin z=\sin w$, then $\sin{\frac{w-z}{2}}\cos{\frac{w+z}{2}}=0$, so either $\frac{w-z}{2}=k\pi$ or $\frac{w+z}{2}=\frac{(2k+1)\pi}{2}$. Since $|Re{(w-z)}|<\frac{\pi}{2}$ in the  (half) strip, the first possibility can hold only for $k=0$, while since $-\pi < Re{(w+z)} <0$  in the (half) strip, the second possibility cannot hold at all, so we must have $z=w$ hence $\sin$ is univalent on the (half) strip (and actually we proved univalence on the closed (half) strip since replacing $<$ with $\le$ gives an extra equality possible only when $w+z=-\pi$ and then as $\Im w,z \ge 0$ we must have $w,z$ real, so $w=z=-\frac{\pi}{2}$).
We now notice the following things: $y \ge 0$ means $\sinh y \ge 0, \cosh y \ge 1$ both in strictly increasing fashion and going to infinity, while $-\frac{\pi}{2} \le x \le 0, y=0$ mean $\sin z$ takes all values between $-1,0$ on the negative real axis.
So looking at the closed (half) strip to which $\sin $ extends continuously (and actually analytically 1-1 too by the above) we know that the image of the closed (half) strip is (the closure of) a simply connected domain with boundary the two axis of the second quadrant by the observation above (the 3 resulting pieces, $x=0, y \ge 0$, $x=-\frac{\pi}{2}, y \ge 0$ and $-\frac{\pi}{2} \le x \le 0, y=0$ cover that completely and without overlap except at the joint ends) so it must be the full closed second quadrant and then obviously (1-1) the inside goes to the inside and we are done
