Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Define

$H^{+}=\{z:y>0\}$

$H^{-}=\{z:y<0\}$

$L^{+}=\{z:x>0\}$

$L^{-}=\{z:x<0\}$

$f(z)=\frac{z}{3z+1}$ maps which portion onto which from above and vice-versa? I will be glad if any one tell me how to handle this type of problem? by inspection?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

HINTS

  1. Fractional transformations/Möbius transformations take circles and lines to circles and lines, i.e. they are 'circilinear.' They also preserve connected regions.
  2. If you find out what happens to the boundaries, you'll know almost everything (except for in which side of the boundary the image resides); in one of those silly word plays, the image of the boundary is the boundary of the image.
  3. Once you know where the upper half plane, say, goes, you know where the lower half plane goes automatically.
share|improve this answer
    
I knew point (1), but how would I apply here that? –  El Angel Exterminador Jun 13 '12 at 11:09
    
@Mex: probably by using points (2) and (3). If the boundary is a line, it's very easy to determine what the image is. In fact, it can be done with just 3 points by brute force. –  mixedmath Jun 13 '12 at 11:14
    
ok in my problem, set 1 has boundary an infinite line. could you please tell me then what I need to check more? –  El Angel Exterminador Jun 13 '12 at 11:18
    
@Mex: Each of your sets is a half-plane. They all have exactly 1 boundary, and they're all lines. In fact, the upper and lower half-planes have the same boundary, as do the 'lefter' and 'righter'. –  mixedmath Jun 13 '12 at 11:22
    
exactly, so they must be mapped onto each other. –  El Angel Exterminador Jun 13 '12 at 11:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.