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Complex Analysis Problem #1

  1. Sep 21, 2007 #1
    1. The problem statement, all variables and given/known data

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

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

    2. The attempt at a solution

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

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

    [tex]\frac{1 - (x^2 + y^2)}{1 - 2y + x^2 + y^2}[/tex].

    Now, I know that the numerator is nonnegative (since [tex] z \in \mathbb{D}, |z| < 1 [/tex], so, [tex] x^2 + y^2 < 1) [/tex]. But, I am not certain about the sign of the denominator in the case where [tex]y > 0[/tex]. 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!
  2. jcsd
  3. Sep 21, 2007 #2


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    This may be complex analysis, but do you still remember (y-1)^2=y^2-2y+1?
  4. Sep 21, 2007 #3
    Man, it's always something really obvious. Thanks.
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