Complex numbers question

• September 21st 2008, 08:30 PM
Johnaloa
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
• September 21st 2008, 08:34 PM
jbpellerin
sorry for spam
• September 22nd 2008, 04:21 AM
mr fantastic
Quote:

Originally Posted by Johnaloa
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
• September 22nd 2008, 04:25 AM
Plato
$\frac{{1 + z}}
{{1 - z}} = \frac{{\left( {1 + z} \right)\left( {1 - \overline z } \right)}}
{{\left| {1 - z} \right|^2 }} = \frac{{1 + z - \overline z - z\overline z }}
{{\left| {1 - z} \right|^2 }} = \frac{{z - \overline z }}
{{\left| {1 - z} \right|^2 }} = \frac{{2yi}}
{{\left| {1 - z} \right|^2 }}$
• September 22nd 2008, 05:13 AM
shawsend
Quote:

Originally Posted by Johnaloa
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.
• September 22nd 2008, 11:24 AM
ThePerfectHacker
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 $|z|=1$), $\arg \left\{ \frac{1+z}{1-z} \right\} = \left\{ \begin{array}{c} \frac{\pi}{2} \text{ if }\Im (z) > 0 \\ -\frac{\pi}{2} \text{ if }\Im (z) < 0 \end{array} \right.$