Mappings in the complex plane

  • Thread starter cragar
  • Start date
  • #1
2,544
2

Homework Statement


What is the mapping of the circle of radius 1 centered at z=-2i
under the mappinf f(z)=1/z

The Attempt at a Solution


I write the circle in polar form [itex] -2i+e^{ix} [/itex]
Now we invert it and multiply by the complex conjugate.
so we get [itex] f(z)= \frac{2i+e^{-ix}}{5+2i(e^{ix}-e^{-ix})} [/itex]
now sure how to graph it our manipulate it any more, should I break it into real and imaginary parts?
also there was a hint that w=1/z and z=1/w. where w is the complex mapping. any help would be appreciated
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
13,599
3,286
Now we invert it and multiply by the complex conjugate
Well, we multiply by 1 where we write 1 as cc/cc. The denominator is now real -- perhaps you want to let it look a little more real (think goniometric functions) ? And for the numerator real and imaginary parts are easily separated...

[edit] I see that there's more to be had here (the image is very simple), but can't pin it down
 
Last edited:
  • #3
2,544
2
thanks for your help, ok when i plug in sin(x) instead of the complex exponentials I get
[itex] \frac{i(2-sin(x)}{5-4sin(x)}+\frac{cos(x)}{5-4sin(x)} [/itex]
Not sure what to do from here, I could plug in points and graph it, but I thought this was to be something somewhat straight forward
to graph.
 
  • #4
BvU
Science Advisor
Homework Helper
2019 Award
13,599
3,286
I could plug in points and graph it
I did just that and it inspired me to the [edit] addition. Just trying to figure it out.
 
  • #5
2,544
2
I guess another way to try and solve it is using the modulus of a circle.
w=1/z. So the radius is one . So we have [itex] |z+2i|=1=|1/w+2i|=1 [/itex]
then we get [itex] |\frac{2iw+1}{w}|=1 [/itex]
then we multiply both sides by |w| and square both sides . Then I get
[itex] 1+4w^2=x^2+y^2 [/itex]
then I write w as x+iy, square it ans then equate the real and imaginary parts.
is this the way to go?
 
  • #6
BvU
Science Advisor
Homework Helper
2019 Award
13,599
3,286
Don't know what the modulus of a circle is. Plug and plot (plod :smile:) gave me another circle as image ...
 
  • #7
2,544
2
ok but how do I get to the circle with algebra ?
 
  • #8
235
62
If ##z=e^{it}## then ##w=\frac{1}{e^{it}}=e^{-it}##. So as ##z## traverses the circle in a counter-clockwise direction, what is ##e^{-it}## doing?
 
  • #9
2,544
2
with the negative sign it goes around the circle in the opposite direction., but when I put the shift in the -2i translation, it makes the algebra more difficult,
my teacher said using the modulus form of circle to make the algebra more clean and simple, but I cant quite get it to work. thanks for your help by the way.
 
  • #10
BvU
Science Advisor
Homework Helper
2019 Award
13,599
3,286
Do you already have the equation of the image circle ?
 
  • #11
2,544
2
I got it figured out, i just need to be more careful with the i and w . thanks for your help.
 

Related Threads on Mappings in the complex plane

Replies
7
Views
6K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
3
Views
631
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
5
Views
1K
Top