Proving $f(x,y)=(e^x \cos y,e^x \sin y)$ Has No Inverse

Question

By the inverse theorem we can see that the Jacobin is $e^x\neq 0$ so there is a locally inverse function, how do I show that it has not inverse function?

Answer 1

In order for a function to have an inverse, it has to be one-to-one.

Notice that For $$ f(x,y)=(e^x \cos y,e^x \sin y)$$

We have $$ f(x,y+2k\pi )=(e^x \cos( y+ 2k\pi) ,e^x \sin( y+2k\pi)) = f(x,y)$$

Thus your function is not one-to-one.