Complex Mapping/Images question

[NewtonianAlch]

Homework Statement

For the mapping f(z) = $\frac{z - i}{z + i}$, find the image of Im(z) ≥ 0

The Attempt at a Solution

So in the z-plane this is clearly everything for y ≥ 0

I substituted x + iy, and rationalised it to get:

$u+iv$ = ${\frac {{x}^{2}-1+{y}^{2}}{{x}^{2}+1+{y}^{2}+2\,y}}-{\frac {2\,ix}{{x}<br /> ^{2}+1+{y}^{2}+2\,y}}$

The denominator can be simplified to x^2 + (y+1)^2

The answer says that this is a closed disc of |w| ≤ 1 with an exclusion at w = 1. Which means I should be seeing an answer of the form u^2 + v^2 ≤ 1

I do not see how they got this.

 

[Dick]

Try looking at what happens to the boundary y=0. Can you show that maps into u^2+v^2=1?

 

[NewtonianAlch]

For (x,y)

I used (0,0) -> (-1,0)
(1,0) -> (0,-1)
(-1,0) -> (0,1)

So one can see that it starts forming a circle, from the answer given I could guess that it is indeed a circle, but since I can't plot (0,-1) since that's not defined, how would I know it is a circle?

Also it's easy to see that for any values y>0, it maps to a point inside the circle, so it's definitely inside the region.

Also, is it possible to see what this is without having to plot several points?

 

[Dick]

For (x,y)

I used (0,0) -> (-1,0)
(1,0) -> (0,-1)
(-1,0) -> (0,1)

So one can see that it starts forming a circle, from the answer given I could guess that it is indeed a circle, but since I can't plot (0,-1) since that's not defined, how would I know it is a circle?

Also it's easy to see that for any values y>0, it maps to a point inside the circle, so it's definitely inside the region.

Also, is it possible to see what this is without having to plot several points?

Put y=0 into your u and v expressions. Show u^2+v^2=1. It's pretty easy algebra.

 

[NewtonianAlch]

Put y=0 into your u and v expressions. Show u^2+v^2=1. It's pretty easy algebra.

I do not quite follow. That gives me an expression in x.

 

[Dick]

I do not quite follow. That gives me an expression in x.

What expression for x? If you did it correctly you should be able to simplify the expression to 1.

 

[NewtonianAlch]

For y = 0:

u + iv = ${\frac {{x}^{2}-2\,ix-1}{{x}^{2}+1}}$

 

[NewtonianAlch]

Oh, I see now, when y=0 and x= 0. Thanks!

 

[Dick]

Oh, I see now, when y=0 and x= 0. Thanks!

Well no, for y=0 and x=anything. ((x^2-1)^2+(2x)^2)/(x^2+1)^2=1. Yes?

 

[NewtonianAlch]

Hmm, yes I see that now after working it out, although I hadn't the faintest idea that would occur by looking at the expressions, how did you figure that out?

 

[Dick]

Hmm, yes I see that now after working it out, although I hadn't the faintest idea that would occur by looking at the expressions, how did you figure that out?

You told me the answer should be u^2+v^2=1. That was my first clue. I know that the function you gave is a Mobius transformation. I know all Mobius transformations map lines to lines or circles. That was a pretty good second clue.

 

[NewtonianAlch]

A Mobius Transformation? Never heard of that, thanks for the tip, I will check it out now.