Complex analysis question

  • #1
131
0

Homework Statement




Let S = {z : 1<= Im(z) <=2}. Determine f(S) if f: S ->C
defined by
f(z) = (z + 1) / (z - 1)





Homework Equations



z = x + iy

The Attempt at a Solution


[attempt at solution]

so here my solution

f(z) = 1 + 2/(z - 1)

after doing some algebra <-> f(z) = x^2 + y^2/((x - 1)^2 + y^2) - [2y/((x-1)^2 + y^2)]i

therefore Im(z) = -2y/((x - 1)^2 + y^2) so F(S) = {z : 1<= (-2y)/((x-1)^2 + y^2) <= 2}
but I am stuck at this point I don't know wat does this represent.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619

Homework Statement




Let S = {z : 1<= Im(z) <=2}. Determine f(S) if f: S ->C
defined by
f(z) = (z + 1) / (z - 1)





Homework Equations



z = x + iy

The Attempt at a Solution


[attempt at solution]

so here my solution

f(z) = 1 + 2/(z - 1)

after doing some algebra <-> f(z) = x^2 + y^2/((x - 1)^2 + y^2) - [2y/((x-1)^2 + y^2)]i

therefore Im(z) = -2y/((x - 1)^2 + y^2) so F(S) = {z : 1<= (-2y)/((x-1)^2 + y^2) <= 2}
but I am stuck at this point I don't know wat does this represent.
Concentrate on what the boundaries of your region are. For example, if 1=(-2y)/((x-1)^2 + y^2) what kind of curve is that? Multiply it out and complete the square. At a more abstract level f(z) is a Mobius transformation. It will map lines to lines or circles, yes?
 
Last edited:
  • #3
131
0
yes I did that I got something weird

I got (x-1)^2 + y^2 <= -2y <= 2( (x - 1)^2 + y^2)) the way I see it its between two circles but how to show that ???
 
  • #4
Dick
Science Advisor
Homework Helper
26,260
619
yes I did that I got something weird

I got (x-1)^2 + y^2 <= -2y <= 2( (x - 1)^2 + y^2)) the way I see it its between two circles but how to show that ???
Just look at the boundaries. Where your inequality becomes an equality. 1=(-2y)/((x-1)^2 + y^2) and 2=(-2y)/((x-1)^2 + y^2). What are the boundary curves? And yes, they are two circles.
 
  • #5
131
0
o I see I figured it out ty alot Dick!
 

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