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

[hadroneater]

Homework Statement

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.

Homework Equations

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.

 

[mighty2000]

Can you give me some more tips?

Thank you for your question. It seems like you are on the right track and have a good understanding of the function f = sin(z) and its properties. To prove that f is one-to-one on the set S, we need to show that for any two points z1 and z2 in S, if sin(z1) = sin(z2), then z1 = z2.

As you have correctly identified, the image of S through f is an ellipse with cosh(y) as the semi-major axis. This means that for any point z in S, its image f(z) also lies on this ellipse. Now, let's assume that there are two points z1 and z2 in S such that sin(z1) = sin(z2). This means that their images f(z1) and f(z2) lie on the same point on the ellipse. Since the ellipse is one-to-one, this can only happen if z1 = z2. Therefore, we can conclude that if sin(z1) = sin(z2), then z1 = z2, and hence f is one-to-one on S.

I hope this helps. Keep up the good work in your studies!