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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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    As @Aaron suggested $f(r,\theta)$ can be obtained by replacing $x$ with $r\cos{\theta}$ and $y$ with $r\sin{\theta}$2012-01-06

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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?2012-01-06