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This is an exercise in Inverse Function Theorem http://en.wikipedia.org/wiki/Inverse_function_theorem

We are given the function $f:\mathbb R^2 \to \mathbb R^2$, $f(x,y)=(e^x \cos y,e^x \sin y)$

  1. Show that $f$ is injective around every point in $\mathbb R^2$ - I managed to solve this.

  2. Find neighborhoods of the points $(0,\pi)$ and $(-1,\frac{\pi}{2})$ such that $f$ is injective in those neighborhoods, and find the inverse function in those neighborhoods.

I managed to do question number 1, but am stumped by question number 2.

The answer to question 1 is that the determinant of the Jacobi matrix is $e^{2x}$ for all $(x,y) \in \mathbb R^2$ and that is always positive and so the determinant in every point on the plane is non-zero, and so we can apply inverse function theorem.

Question number 2...Could use a hand

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  • $\begingroup$ I take it that with environments you mean neighborhoods. $\endgroup$
    – Git Gud
    Commented Mar 22, 2014 at 15:02
  • $\begingroup$ Yes thank you...English is not my native tongue. I do apologize. $\endgroup$ Commented Mar 22, 2014 at 15:02
  • $\begingroup$ If you can solve 1., then 2. follows easily. In 1. you proved that $f$ is injective around every point, so take as a neighborhood for 2. one of the neighborhoods in which you proved that $f$ is injective in 1. $\endgroup$
    – Git Gud
    Commented Mar 22, 2014 at 15:21
  • $\begingroup$ But I did not find such neighborhood. I just said "There's a neighborhood around every point". I don't know how to find said neighborhood. $\endgroup$ Commented Mar 22, 2014 at 15:22
  • $\begingroup$ How did you prove such a neighborhood exists? $\endgroup$
    – Git Gud
    Commented Mar 22, 2014 at 15:23

1 Answer 1

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Suppose we write \begin{cases}u&=&e^x\cos y\\ v&=&e^x\sin y\end{cases} so that $f(x,y)=(u,v)$.

Can you solve that system of equations for x and y? If so, what are the conditions for the solution to be a valid inverse around the points you mentioned?

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    $\begingroup$ I get $x=\frac{1}{2}ln(u^2+v^2)$ and $y=arccos(\frac{u}{\sqrt{u^2+v^2}})$ is this correct? How does this help? $\endgroup$ Commented Mar 22, 2014 at 15:36
  • $\begingroup$ Good! However, it should be $y=\pm\arccos\left(\frac u {\sqrt{u^2+v^2}}\right) + 2k\pi \ (k \in \mathbb Z)$. As you can see, it is not defined for u=v=0 only, and furthermore there are multiple solutions for y. For your problem you have to selection 1 of those solutions and pick neighborhoods that in particular exclude (u,v)=(0,0). $\endgroup$ Commented Mar 22, 2014 at 15:41
  • $\begingroup$ And how will we find the inverse? $\endgroup$ Commented Mar 22, 2014 at 15:47
  • $\begingroup$ Huh? This is the inverse. $\endgroup$ Commented Mar 22, 2014 at 15:49
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    $\begingroup$ That is ofcourse correct. Thank you friend. $\endgroup$ Commented Mar 22, 2014 at 15:53

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