Complex analysis. Inversion in the complex plane?? (Proofs)

Question

anonymous · Answer

I'm not sute if the same ideas apply to the regular xy-plane/euclidean geometry, but I need to somehow prove these properties of inversion:

The image of a circle that goes through the center is a straight line that doesn't go through the center

The image of a straight line that doesn't go through the center is a circle that goes through the center

The image of a circle that doesn't go through the center is another circle that doesn't go through the center

The image of a straight line that goes through the center is the same straight line


No idea how to even start with these. I mean, I have random equations mentioned in class, but I'm not sure how to tie them into all of this.

anonymous · Answer

I guess the equation for inversion is 
$$w = \frac{ 1 }{\overline{z}  } = u + iv$$

And then since $\overline{z} = x - iy$, you can say $x = \frac{u}{u^{2}+v^{2}}$, $y = \frac{v}{u^{2}+v^{2}}$. How to use all f this, though, I have no clue.

usukidoll · Answer

go to math.stackexchange.com  you will  have better luck having an answer. trust me.

anonymous · Answer

Still is always worth a try to post here as well.

zzr0ck3r · Answer

I don't know if I have seen complex analysis on here. Real analysis and topology and even convex analysis, but not complex. Maybe @eliassaab

anonymous · Answer

Ah. There are just so many brilliant people on here that I figure it's always worth a try to see what someone may know

anonymous · Answer

Let me prove this: The image of a straight line that goes through the center is the same straight line.

Let us take the line  v = a u and see what is its image

$$
x=\frac{u}{u^2+v^2}=\frac{u}{a^2
   u^2+u^2}=\frac{1}{a^2 u+u} \
   
 y=\frac{v}{u^2+v^2}=\frac{a u}{a^2
   u^2+u^2}=\frac{a}{a^2 u+u}=a x  


$$
Try to do the rest using the same technique

anonymous · Answer

@Concentrationalizing

anonymous · Answer

@zzr0ck3r

anonymous · Answer

Let me do this
The image of a straight line that doesn't go through the center is a circle that goes through the center
Take the line $$v = a u + b; \quad b\ne 0
$$
Then 
$$
x=\frac{u}{(a u+b)^2+u^2} \
y=\frac{a u+b}{(a u+b)^2+u^2}
$$
Eliminate u to get $$ y (1-b y)=a x+b x^2$$ which is a circle containing (0,0)

anonymous · Answer

@UsukiDoll

anonymous · Answer

See also this http://www.math.bas.bg/~rkovach/lectures/Moebius.pdf

zzr0ck3r · Answer

I hope he came back...

usukidoll · Answer

How the heck would I know if i haven't studied this subject?