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So, I have $ f(x,y) = (x^2-y^2, 2xy) $, which is a local $\mathcal C^1$ isomorphism in $\mathbb R^2 \setminus \{(0,0)\}$.

I have to write this function in polar coordinates:

$$f(x,y) = f(r\cos\phi, r\sin\phi).$$

My beginnings: I know that

$$df(r, \phi) = \cos\phi -r\sin\phi \sin\phi r\cos\phi.$$

But I really have no clue how to work this one out. It is a single exercise of this kind in my book; so, I probably don't need it for the test, but I would like to know.

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Just plug in $r\cos \phi$ and $r\sin\phi$ for $x$ and $y$ respectively. Note that trig identities will allow you to simplify your result. –  Aaron Jan 6 '12 at 10:48
    
As @Aaron suggested $f(r,\theta)$ can be obtained by replacing $x$ with $r\cos{\theta}$ and $y$ with $r\sin{\theta}$ –  Ramana Venkata Jan 6 '12 at 10:59

2 Answers 2

In polar coordinates, we have $(x,y) =(r\cos\phi, r\sin\phi)$ as you said. Therefore, $f(x,y) = (x^2-y^2, 2xy)$ can be rewritten as $$f(x,y) =f(r\cos\phi, r\sin\phi)= \big((r\cos\phi)^2-(r\sin\phi)^2, 2(r\cos\phi)(r\sin\phi)\big)$$ $$=\big(r^2(\cos^2\phi-\sin^2\phi),r^2(2\sin\phi\cos\phi)\big)=\big(r^2\cos(2\phi),r^2\sin(2\phi)\big).$$

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thanx much paul, i came to similar solution. Would that mean that the function is bijective only from (0,8)x(0, 2Pi) because of ϕ being periodic? –  dubiousa Jan 6 '12 at 17:49

You don't say what class this is for. Is there any chance this is a complex variables class? If so, you might notice that the $x$ and $y$ components of your formula are the real and imaginary components of $z^2 = (x+i y)^2$. More generally, this approach can work if your transformation satisfies the Cauchy-Riemann equations.

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its multidimensional analysis, and C(p) isomorphism. Like if u have f: U to V, if there is an f^(-1): V to U, if and where. –  dubiousa Jan 6 '12 at 17:45

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