# Mappings in the complex plane

1. Feb 9, 2017

### cragar

1. The problem statement, all variables and given/known data
What is the mapping of the circle of radius 1 centered at z=-2i
under the mappinf f(z)=1/z
3. The attempt at a solution
I write the circle in polar form $-2i+e^{ix}$
Now we invert it and multiply by the complex conjugate.
so we get $f(z)= \frac{2i+e^{-ix}}{5+2i(e^{ix}-e^{-ix})}$
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

2. Feb 9, 2017

### BvU

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...

 I see that there's more to be had here (the image is very simple), but can't pin it down

Last edited: Feb 9, 2017
3. Feb 9, 2017

### cragar

thanks for your help, ok when i plug in sin(x) instead of the complex exponentials I get
$\frac{i(2-sin(x)}{5-4sin(x)}+\frac{cos(x)}{5-4sin(x)}$
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. Feb 9, 2017

### BvU

I did just that and it inspired me to the  addition. Just trying to figure it out.

5. Feb 10, 2017

### cragar

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 $|z+2i|=1=|1/w+2i|=1$
then we get $|\frac{2iw+1}{w}|=1$
then we multiply both sides by |w| and square both sides . Then I get
$1+4w^2=x^2+y^2$
then I write w as x+iy, square it ans then equate the real and imaginary parts.
is this the way to go?

6. Feb 10, 2017

### BvU

Don't know what the modulus of a circle is. Plug and plot (plod ) gave me another circle as image ...

7. Feb 10, 2017

### cragar

ok but how do I get to the circle with algebra ?

8. Feb 11, 2017

### aheight

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. Feb 11, 2017

### cragar

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. Feb 11, 2017

### BvU

Do you already have the equation of the image circle ?

11. Feb 11, 2017

### cragar

I got it figured out, i just need to be more careful with the i and w . thanks for your help.