Prove function to be one-to-one: Complex Mapping

1. Feb 2, 2012

1. The problem statement, all variables and given/known data

Show that f = sin(z) is one-to-one in the set S = { z | -pi < Re(z) < pi, Im(y) > 0}. That is, show that if z1 and z2 are in S and sin(z1) = sin(z2) then z1 = z2.

Hint: Try to find the image of S through f.

2. Relevant equations

3. The attempt at a solution
z = x + iy
sin(z) = i/2(e^-iz - e^iz) = sin(x)cosh(y) + icos(x)sinh(y) = u + i*v

I try to map the function.

know: sin(x)^2 + cos(x)^2 = 1 = u^2/cosh(y)^2 + v^2/sinh(y)^2
If I fix y = constant > 0, I get ellipses with cosh(y) as the semi-major axis. The ellipse becomes a near-circle as y becomes larger.

know: cosh(y)^2 - sinh(y)^2 = 1 = u^2/sin(x)^2 - v^2/cos(x)^2
If I fix x = constant between pi and -pi, I get a hyperbola. If x = pi or -pi, I get the imaginary axis.

I was able to draw the image. However...I am not sure how I can use it to prove f is one-to-one on S.