# Complex Analysis Problem #1

1. Sep 21, 2007

### BSMSMSTMSPHD

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

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?

2. Sep 21, 2007

### Dick

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

3. Sep 21, 2007

### BSMSMSTMSPHD

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