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### Transformations in the Complex Plane

```
Date: 12/07/98 at 10:53:38
From: Tootsie
Subject: Transformations in the Complex Plane

Dr. Math,

I have 2 questions about complex transformations:

1) I am supposed to find the set in the complex plane defined by the
condition Im(1/z) < -1/2, and tell if the set is open, connected, and
bounded.

2) I need to show that the transformation w = 2/z maps the disk
abs(z-i) < 1 onto the lower half of the plane Im(w) < -1. What does
this look like on the Riemann sphere? I think that the 2/z should be
like the 1/z somehow, but I'm not sure.

I understand open, connected, and bounded, but I could really use your
help.  Thanks.
```

```
Date: 12/07/98 at 16:22:28
From: Doctor Anthony
Subject: Re: Transformations in the Complex Plane

Question 1:

If z = x+iy:

1      (x-iy)      x-iy
1/z =  ------ * ------  = -------
x+iy    (x-iy)    x^2+y^2

and so we require:

-y
-------- < -1/2
x^2+y^2

-2y < -(x^2+y^2)

x^2 + y^2 - 2y < 0

x^2 + y^2 - 2y + 1 < 1

x^2 + (y-1)^2 < 1

So the region in the plane satisfying the inequality is the interior
of the circle with centre (0,1) and radius 1. The set is bounded and
connected.

Question 2:

If w = 2/z then z = 2/w. So |z-i| < 1 can be written  |2/w -i| < 1,
and 2/w = 2/(u+iv) = 2(u-iv)/(u^2+v^2) = 2w*/|w|^2.

So we have |2w*/|w|^2 - i| < 1, and using the fact that |z|^2 = z.z* we
have on squaring the above inequality:

(2w*/|w|^2 - i)(2w/|w|^2 + i) < 1

4ww*/|w|^4 + i(2w*/|w|^2 - 2w/|w|^2) + 1 < 1

4|w|^2/|w|^4 + 4v/|w|^2  < 0

1/|w|^2 + v/|w|^2 < 0

1 + v < 0

v < -1

That is, Im(w) < -1.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Imaginary/Complex Numbers
High School Imaginary/Complex Numbers

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