Prove that the map $f(x,y) = (e^x\cos y,e^x\sin y)$ is not injective in the plane

Question

Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $f(x,y) = (e^x\cos y,e^x\sin y)$

a) Prove that if $\det f'(x,y) \neq 0$ $\forall (x,y) \in \mathbb{R}^2$.

b) Show that for any $u = (x,y) \in \mathbb{R}^2$ there exists a $\delta > 0$ such that $f$ is invertible in $B(u,\delta)$.

c) Prove that $f$ is not injective.

So, I've already proven a, and I can use that, given that the partial derivatives are continuous they are derivable, so $f$ is a at least $C^1$ (one time derivable) so there exists a $B(u,\delta)$ where the function is injective then it's invertible (I'm not sure with that last part, I'm having trouble there, any hints?)

For c, I guess I've to show that as it may be injective in a $B(u,\delta)$ it is not injective for $f's$ whole domain but I have no clue there.

Answer 1

Note that $f(x,y) = f(x, y+2\pi)$ for all $x,y \in \Bbb{R}$; hence $f$ is not injective. This solves (c).

For (b) directly check whether $f$ satisfies the hypotheses of the inverse function theorem (clearly $f$ does).

Answer 2

for b) use the Inverse Function Theorem. For c), just find two points in $\mathbb{R}^2$ that get mapped to the same point. For simplicity, let $x = 0$. Then $f(0,y) = \left(cos(y), sin(y)\right)$. Can you think of two different y values that get mapped to the same point? Look at the unit circle.