Proving the One-to-One Property and Image of a Complex Function

[BSMSMSTMSPHD]

Homework Statement

Let f(z) = \frac{1-iz}{1+iz} and let \mathbb{D} = \{z : |z| < 1 \}.

Prove that f is a one-to-one function and f(\mathbb{D}) = \{w : Re(w) > 0 \}.

2. The attempt at a solution

I've already shown the first part: Assume f(z_1) = f(z_2) for some z_1, z_2 \in \mathbb{C}, then z_1 = z_2. (I worked this out).

But for the second part, I'm not sure what to do. I've written the function in rectangular coordinates (z = x + iy) and the real part of the simplified fraction is:

\frac{1 - (x^2 + y^2)}{1 - 2y + x^2 + y^2}.

Now, I know that the numerator is nonnegative (since z \in \mathbb{D}, |z| < 1, so, x^2 + y^2 < 1). But, I am not certain about the sign of the denominator in the case where y > 0. Any ideas? And, if I can show this, will I have finished the proof, or do I have to show reverse inclusion?

Thanks in advance!

 

[Dick]

This may be complex analysis, but do you still remember (y-1)^2=y^2-2y+1?

 

[BSMSMSTMSPHD]

Man, it's always something really obvious. Thanks.