i totally misread that
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Hey guys, I cant figure this out :
z is a complex number for which
Show that
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
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:
. Try and recognize quotients like these as Mobius or linear fractional transformation: circles to lines and circles. But for now:
First solve for z: . Now, if then or . 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 . You can see this more clearly if you express the absolute value quotient in terms of and then complete the squares. The and terms cancel and you're left with a line.
The mapping is a Mobius transformation as shawsend mentioned.
The circle (minus the point ) 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 then a simple computation shows it gets mapped to . Therefore, (minus the point ) is mapped into the line . And it follows if you pick any point on this line then its argument is either . Thus, what you wrote it not entire correct. A little more work will reveal the the top portion of the circle gets mapped into and bottom portion of the circle gets mapped into .
Therefore (for and ),