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Math Help - Complex numbers question

  1. #1
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    Complex numbers question

    Hey guys, I cant figure this out :

    z is a complex number for which  |z| = 1

    Show that  \arg\left|\frac{1+z}{1-z}\right| = \frac{\pi}{2}

    Can somebody show me how to do this. What I did was to take z = x + iy and then input it into the equation. Rationalised it and then ended up getting x = iy. So z = 2iy. But I still dont get how to solve it? I get that 1+z/1-z has to be totally imaginary for it to have an argument of pi/2.

    But with my method it doesnt turn out to be totally imaginary. Its 1+2iy/1-2iy which still has that real component. What am I doing wrong? Am I even on the right track?

    Thanks in advance for the help
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  2. #2
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    i totally misread that
    sorry for spam
    Last edited by jbpellerin; September 21st 2008 at 07:35 PM. Reason: woah
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  3. #3
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    Quote Originally Posted by Johnaloa View Post
    Hey guys, I cant figure this out :

    z is a complex number for which  |z| = 1

    Show that  \arg\left|\frac{1+z}{1-z}\right| = \frac{\pi}{2}

    Can somebody show me how to do this. What I did was to take z = x + iy and then input it into the equation. Rationalised it and then ended up getting x = iy. So z = 2iy. But I still dont get how to solve it? I get that 1+z/1-z has to be totally imaginary for it to have an argument of pi/2.

    But with my method it doesnt turn out to be totally imaginary. Its 1+2iy/1-2iy which still has that real component. What am I doing wrong? Am I even on the right track?

    Thanks in advance for the help
    Believe it or not, there's an elegant solution involving the circle theorem discussed at this link: http://www.math.toronto.edu/barbeau/diamangleproof.pdf
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  4. #4
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    \frac{{1 + z}}<br />
{{1 - z}} = \frac{{\left( {1 + z} \right)\left( {1 - \overline z } \right)}}<br />
{{\left| {1 - z} \right|^2 }} = \frac{{1 + z - \overline z  - z\overline z }}<br />
{{\left| {1 - z} \right|^2 }} = \frac{{z - \overline z }}<br />
{{\left| {1 - z} \right|^2 }} = \frac{{2yi}}<br />
{{\left| {1 - z} \right|^2 }}
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  5. #5
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    Quote Originally Posted by Johnaloa View Post
    Hey guys, I cant figure this out :

    z is a complex number for which  |z| = 1

    Show that  \arg\left|\frac{1+z}{1-z}\right| = \frac{\pi}{2}
    Hey John, that's not the right notation I think: The argument of the absolute value is zero right? It's a real number. I think you mean:

    \text{Arg}\left\{\frac{1+z}{1-z}\right\}. Try and recognize quotients like these as Mobius or linear fractional transformation: circles to lines and circles. But for now:

    First solve for z: w=\frac{-(1-w)}{1+w}. Now, if |z|=1 then 1=\frac{|1-w|}{|1+w|} or |1+w|=|1-w|. So that the distance of w from 1 is the same as the distance from -1. That's the perpendicular bisector of the two points which is the imaginary axis, that is, the argument is \pi/2. You can see this more clearly if you express the absolute value quotient in terms of w=x+iy and then complete the squares. The x^2 and y^2 terms cancel and you're left with a line.
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  6. #6
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    The mapping z\mapsto \frac{1+z}{1-z} is a Mobius transformation as shawsend mentioned.
    The circle |z| = 1 (minus the point z=1) is therefore mapped into a line through the origin.

    Since a line is determined by two points and the origin is one of the points it means we just need to determine another point. Let us choose z=i then a simple computation shows it gets mapped to i. Therefore, |z|=1 (minus the point z=1) is mapped into the line \{ z \in \mathbb{C} : \Re (z) = 0\}. And it follows if you pick any point on this line then its argument is either \pm \frac{\pi}{2}. Thus, what you wrote it not entire correct. A little more work will reveal the the top portion of the circle gets mapped into \{ z\in \mathbb{C} : \Re (z) = 0 \text{ and }\Im (z) > 0 \} and bottom portion of the circle gets mapped into \{ z\in \mathbb{C} : \Re (z) = 0 \text{ and }\Im (z) < 0 \}.

    Therefore (for z\not = \pm 1 and |z|=1), \arg \left\{ \frac{1+z}{1-z} \right\} = \left\{ \begin{array}{c} \frac{\pi}{2} \text{ if }\Im (z) > 0 \\ -\frac{\pi}{2} \text{ if }\Im (z) < 0 \end{array} \right.
    Last edited by ThePerfectHacker; September 22nd 2008 at 07:49 PM.
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