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/
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