# Finding the image of a set under f?

1. Oct 28, 2010

### Buri

1. The problem statement, all variables and given/known data

Well I'm looking for a 'general' way about doing these problems. For example finding B = f(A) where f(x,y) = (x² - y², 2xy) and A consisting of all (x,y) with x > 0.

The attempt at a solution

I've tried considering the boundary, but I find that it just gets rather confusing. Is there a simpler way of doing problems like this? Btw, how would considering the boundary even work? Couldn't the image go really crazy? But continuity and differentiability probably guarantee the function is pretty well behaved eh?

Thanks for the help!

2. Oct 28, 2010

### Dick

Hmm. Same question came up just a day or two ago. It is a special function. If you know complex variables it's the image of the complex function f(z)=z^2, where z=x+iy. If you don't then just write it in polar coordinates. But thinking about the boundary isn't a bad idea either.

Last edited: Oct 28, 2010
3. Oct 28, 2010

### Buri

I've actually never worked with polar coordinates before, but I read about it just now on wiki and it would seem that (x² - y², 2xy) would become (r²cos(2θ), r²sin(2θ)), right? So its basically considering now when x = rcos(θ) > 0? So on the interval (-π/2, π/2)? Actually, why not on a different interval?

EDIT: Wait so I suppose now that I must look at -π < 2θ < π? But cos(2θ) and sin(2θ) take on all values between (-1, 1) on this interval, but r can take on any value, so it includes everything in the plane?

Last edited: Oct 28, 2010
4. Oct 28, 2010

### Buri

Okay trying to do this with the boundary I have:

If x is equal to 0, we would obtain (-y², 0). So I guess the negative x axis is out and so will the origin. But this is where I get confused. How do I know that these will be the only points not in B? I guess I'd like to show that any point in R² can be written as (x² - y², 2xy), but I just can't do that...

5. Oct 28, 2010

### Buri

Well I'm getting different answers, so I must be doing something wrong...any help?

6. Oct 29, 2010

### Dick

No, you aren't getting different answers. The answer is the real plane minus the negative real axis and the origin. The negative real axis is theta=pi. The image corresponds to the angular range -pi<theta<pi. theta=pi isn't in there.

7. Oct 29, 2010

### Buri

How would I know that everything else is actually included? Is checking the boundary really sufficient? I guess that's what I'm having troubles with..

8. Oct 29, 2010

### Buri

Like I can't see a way of actually getting it algebraically that everything else is inside, or at least 'see' it.

9. Oct 29, 2010

### Dick

No, I don't think knowing where the boundary of a region is mapped is going to tell you everything about where the region is mapped. But it can be a good clue. The whole point to doing it in polar coordinates was to be able to 'see' it. If (r,theta) is a point in f(A) with -pi<theta<pi, then the corresponding point in A is (sqrt(r),theta/2). You could write down the corresponding expressions in x-y coordinates, but it would be complicated and you wouldn't be able to 'see' it.

10. Oct 29, 2010

### Buri

f(x,y) = ((e^x)cosy, (e^x)siny) for A (x,y) such that 0 < y < 2π.

I considered the boundary once again. If y = 0 then f(x,0) = (e^x, 0). So the positive x axis is excluded (not including the origin since e^x > 0 for all x). And now if y = 2π we get the same thing. But is this all that's excluded from R²? See I considered looking at what points the function would map if y = π and I get f(x,π) = (-e^x, 0) so this is the negative axis. But if I had taken a look at y = -π I would get the exact same thing, but this y isn't in the set A. So shouldn't it also be excluded? Now that I think about it, can't this function not map to (0,0)?

11. Oct 29, 2010

### Buri

Thanks a lot for your help btw.

12. Oct 29, 2010

### Dick

Everything you've said so far is fine, except suggesting that the negative axis should be excluded because it's mapped to by y=(-pi) as well as y=pi. Just because something in f(A) is also the image of something outside of A doesn't mean you exclude it.