# Mobius transformation for the first quadrant

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1. Sep 5, 2016

### Zeeree

1. The problem statement, all variables and given/known data
Find the images of the following region in the z-plane onto the w-plane under the linear fractional transformations

The first quadrant $x > 0, y > 0$ where $T(z) = \frac { z -i } { z + i }$

2. Relevant equations

3. The attempt at a solution

So for this, I looked at the poles of $T(z)$ first and found that $z = -i$ does not lie on the lines that bound the first quadrant i.e $x = 0$ and $y = 0$. Since the image of a line is either a line or a circle, I deduced that the image is a circle since the singularity does not lie on the lines.

Upon substituting $z = 0$ and $z = 1$ where both are points on the bounding lines, I obtained $T(0) = -1$ and $T(1) = -i$ leading me to believe that it in fact the unit circle (Exterior or interior can be found out later)

However, because it's a mapping of only the FIRST quadrant, intuitively I think the image is the semi-circle but I'm unsure how to show this. If I were to substitute $z = -1$ into T(z), I obtain $T(z) = i$ which gives me the hunch that it's the upper half unit circle (since $z = -1$ is not in the first quadrant. I don't think this is enough.

2. Sep 5, 2016

### pasmith

Show:
(1) If $x > 0$ then $|T(x)| = 1$ and $T(x)$ lies in the lower half plane.
(2) If $y > 0$ then $-1 < T(iy) < 1$.
(3) T(1 + i) lies in the lower half plane and $|T(1 + i)| < 1$.

3. Sep 6, 2016

### Zeeree

For the first one, I've shown that $T(x)$ (for any value of $x > 0$) takes the form $\frac { x^2 - 1 } { x^2 + 1 } - \frac { 2x } { x^2 + 1 } i$, meaning the v-value in the w-plane will always be negative for any $x > 0$

I'm not so sure about the 2nd one. $T(iy)$ yields $\frac { i(y - 1) } { i(y + 1) }$. Not sure how to proceed after canceling the $i$s

However, I've managed to do the third one. Thanks!