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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?

2 Answers 2

1

$f(z)=\frac{z}{3z+1}=\frac{x+iy}{3(x+iy)+1}=\frac{x+iy}{3x+1+i3y}=\frac{(x+iy)(3x+1-i3y)}{(3x+1)^2+9y^2} \implies \Im (f(z))=\frac{y}{(3x+1)^2+9y^2}$. So, If $y > 0$, then $\Im (f(z))>0$.If $y < 0$, then $\Im (f(z))<0.$

2

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.
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    I knew point (1), but how would I apply here that?2012-06-13
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    @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.2012-06-13
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    ok in my problem, set 1 has boundary an infinite line. could you please tell me then what I need to check more?2012-06-13
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    @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'.2012-06-13
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    exactly, so they must be mapped onto each other.2012-06-13
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    @Mex: if you are saying that because two things share the same boundary, every fractilinear transormation takes one onto the other, then you are incorrect. I suggest you literally find the image of the boundaries.2012-06-13