Exploring the Set f(A) of the Function f(x,y)

[JG89]

Homework Statement

Let $$f(x,y) = (x^2 - y^2, 2xy)$$ where f is defined on the open set $$A = \{ (x,y) \in \mathbb{R}^2 : x > 0 \}$$.

What is the set f(A)?

Homework Equations

The Attempt at a Solution

I'm guessing they're asking me to describe f(A) using set notation. But I don't know how to do this. I've tried plotting a set of points to get an idea what f(A) would look like on the plane, but I can't get any idea whatsoever.

Any help or ideas?

 

[Dick]

If you think of (x,y) as the complex variable z=x+iy, then it's the image of the half plane x>0 under mapping f(z)=z^2. Think about expressing it in polar form.

 

[JG89]

Look at below post...

 

[JG89]

Dick, I'm not too familiar with polar coordinates so please bear with me.

We can write, $$f(z) = (x + iy)^2 = (x^2 - y^2) + i(2xy)$$. Using polar coordinates, we have $$r = \sqrt{ (x^2 - y^2)^2 + (2xy)^2 } = \sqrt{ x^4 - 2x^2 y^2 + y^4 + 4x^2y^2} = \sqrt{ x^4 + 2x^2y^2 + y^4} = \sqrt{(x^2 + y^2)^2} = x^2 + y^2.$$

So, we can write $$f(z) = (x^2 - y^2) + i(2xy) = (r^2 - 2y^2) + i(2xy)$$. Since $$x = r cos(t), y = r sin(t)$$ then $$f(z) = (r^2 - 2y^2) + i(2xy) = r^2 - 2r^2 sin^2 (t) + i(2r^2sin(t) cos(t)) = r^2 [ 1 - 2sin^2 (t) + i (2 sin(t) cos(t))] = r^2[ cos(2t) + i(sin(2t)]$$. I used a bunch of trig identities for that...

Anyway, so we see that $$f(z) = f(x + iy) = f( r[cos(t) + isin(t)]) = r^2[ cos(2t) + i sin(2t)]$$.

So the distance from the origin to the point f(z) is r^2 (whereas the distance from the origin to the point z is r) and the polar-angle is 2t, twice the original angle. Now I have a better idea of how f(A) would look on the plane, but I still don't know how to express it with set notation, or in a way that seems appropriate to answer the question.

 

[Dick]

You are making this way too complicated. And it's also wrong. z=x+iy, r=sqrt(x^2+y^2). If x>0 then z=r*exp(i*theta) where theta is in (-pi/2,pi/2), right? You do know something about complex variables, yes? I was hoping so.

 

[JG89]

Ok, so $$f(z) = f(r e^{i \theta}) = (r e^{i \theta})^2 = r^2 e^{i(2 \theta)}$$

I still don't see anything...

EDIT: I know a bit, but maybe not enough for this question, as I think I might have made an exponent mistake just now because of the complex variables [/tex]

 

[Dick]

It's simpler than you think. You were setting r equal to |f(z)|. r should be |z| to use the expression you are suggesting. Do you agree -pi/2<theta<pi/2?

 

[JG89]

f(z) = r^2 exp(i*2*theta) = r^2[ cos(2*theta) + i sin(2*theta) ].

Again it just seems to me that the distance from the origin to f(z) is the square of the distance from the origin to z, and that the angle between the positive x-axis and f(z) is double the angle from the positive x-axis to z. Just the same stuff I said before...

Maybe I need to catch some sleep then think about it in the morning, or am I onto something?...EDIT: Yes, I agree that -pi/2 < theta < pi/2, which would mean that -pi < 2*theta < pi...

 

[Dick]

f(z) = r^2 exp(i*2*theta) = r^2[ cos(2*theta) + i sin(2*theta) ].

Again it just seems to me that the distance from the origin to f(z) is the square of the distance from the origin to z, and that the angle between the positive x-axis and f(z) is double the angle from the positive x-axis to z. Just the same stuff I said before...

Maybe I need to catch some sleep then think about it in the morning, or am I onto something?...


EDIT: Yes, I agree that -pi/2 < theta < pi/2, which would mean that -pi < 2*theta < pi...

Yeah, you are just picturing this wrong somehow. f(A) is ALL points where r>0 and -pi<theta<pi. ALL of them. f(A) isn't some kind of curve.

 

[JG89]

So f(A) is just the plane, minus the origin and the negative x-axis?

 

[Dick]

So f(A) is just the plane, minus the origin and the negative x-axis?

That's my conclusion.

 

[JG89]

Ahhh, I see it now. Thanks for the help. I have a similar question just with a different map, so I'm going to go work on that now. I should be fine with it :)