# Inverse function theorem question - multivariable calculus

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

• I take it that with environments you mean neighborhoods. Commented Mar 22, 2014 at 15:02
• Yes thank you...English is not my native tongue. I do apologize. Commented Mar 22, 2014 at 15:02
• 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. Commented Mar 22, 2014 at 15:21
• 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. Commented Mar 22, 2014 at 15:22
• How did you prove such a neighborhood exists? Commented Mar 22, 2014 at 15:23

Suppose we write \begin{cases}u&=&e^x\cos y\\ v&=&e^x\sin y\end{cases} so that $f(x,y)=(u,v)$.
• 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? Commented Mar 22, 2014 at 15:36
• 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). Commented Mar 22, 2014 at 15:41