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/