Mobius transformation for the first quadrant

[Zeeree]

Homework Statement

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 }##

Homework Equations

The Attempt at a Solution

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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.

 

[pasmith]

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

 

[Zeeree]

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

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!