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Functions composition

  • Thread starter skrat
  • Start date
  • #1
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Homework Statement


Find conformal function that maps ##D_1=\left \{ z;Re(z)>0,0<Im(z)<2\pi \right \}## on to ##D_2## where ##D_2## is unit disc.


Homework Equations





The Attempt at a Solution



Ok, I haven't got any problems with conformal mapping but I have huge problems with function composition and I would need some help here...

Firstly, ##f_1=e^z## maps from ##D_1## into unit disc without ##Re(z)>0## axis, so to this object I now apply ##f_2=\sqrt{z} ## which gives me upper half of unit disc (real exis not included!).

Now ##f_3=\frac{z+1}{1-z}## maps the upper unit disk into first quadrant. Applying ##f_4=z^2## extends my area on to upper half plane. Rotating it with ##f_5=ze^{-i\pi /2}## and again using Mobius transformation ##f_6=\frac{i-z}{z+i}## gives me that unit disk.

Now the question is of course what ##f=f_6\circ f_5\circ f_4\circ f_3\circ f_2\circ f_1##?

Does it even make any sense to do that?
 

Answers and Replies

  • #2
Fredrik
Staff Emeritus
Science Advisor
Gold Member
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Homework Statement


Find conformal function that maps ##D_1=\left \{ z;Re(z)>0,0<Im(z)<2\pi \right \}## on to ##D_2## where ##D_2## is unit disc.


Homework Equations





The Attempt at a Solution



Ok, I haven't got any problems with conformal mapping but I have huge problems with function composition and I would need some help here...

Firstly, ##f_1=e^z## maps from ##D_1## into unit disc without ##Re(z)>0## axis,
##f_1(1+i\pi)=e^{1+i\pi}=e^1 e^{i\pi}= -e## is not in the unit disc.
 
  • #3
746
8
Of course not. It even doesn't have to be.

There is a typo in my original post. Instead of ##D_1=\left \{ z;Re(z)>0,0<Im(z)<2\pi \right \}## it is ##D_1=\left \{ z;Re(z)<0,0<Im(z)<2\pi \right \}##
 

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