Complex Variables. Problem about complex sine.

[ELESSAR TELKONT]

[SOLVED] Complex Variables. Problem about complex sine.

Homework Statement

Proof that the function
$$\begin{displaymath}<br /> \begin{array}{cccc}<br /> f: &A=\left\{z\in\mathbb{C}\mid-\frac{\pi}{2}<\Re z<\frac{\pi}{2}\right\} &\longrightarrow &B=\mathbb{C}-\left\{z\in\mathbb{C}\mid \Im{z}=0,\,\left\vert\Re z\right\vert\geq 1\right\}\\<br /> &z &\mapsto &\sin(z)<br /> \end{array}<br /> \end{displaymath}$$
is a biyection.

Homework Equations

The Attempt at a Solution

I have already proven (on the previous exercise of my book) that the complex sine is inyective on a vertical band of the complex plane of width $$\pi$$. Then I have only to proof that the function is surjective or that there exists left inverse (more difficult since its multivaluated in general).

My problem is that I have no idea how to proof that since I have saw that the complex sine maps vertical lines in the complex plane to hyperbolas and horizontal ones to ellipses. Then I think it's imposible that the sine could map the set $$A$$ to the set $$B$$ one-one because the hyperbolas and the ellipses I have mentioned get out from $$B$$.

 

[e(ho0n3]

I did this very same problem recently. Fortunately, the book had already proven that sin(z) maps the strip $\{x + iy : 0 \le x \le \pi/2$ and $y \ge 0\}$ in a one-to-one fashion onto the first quadrant.

Then, using the fact that $\sin(-\bar{z}) = -\overline{\sin(z)}$, one may show that sin(z) maps the strip $\{x + iy : -\pi/2 \le x \le \pi/2$ and $y > 0\}$ to the upper half-plane bijectively.

Similarly, one may show that sin(z) maps the strip $\{x + iy : -\pi/2 \le x \le \pi/2$ and $y < 0\}$ to the lower half-plane bijectively.

Now all that's left is to consider the strip $\{x + iy : -\pi/2 \le x \le \pi/2$ and $y = 0\}$ which I leave to you.

 

[ELESSAR TELKONT]

Thanks for your help