Range and injectivity of f(x,y) = (e^x cos y, e^x sin y)

[linearfish]

Homework Statement

Let f = (f₁,f₂) be the mapping of R² into R² given by f₁=e^xcos(y), f₂=e^xsin(y).
(1) What is the range of f?
(2) Show that every point of R² has a neighborhood in which f is one-to-one.
(3) Show that f is not one-to-one on R².

The Attempt at a Solution

(1) Since e^x is nonzero and cos and sin cannot be 0 at the same time, then the origin is not in the range, so the range is R²\{0,0}.

(2) Calculating the Jacobian we get:
J = e^2xcos²y + e^2xsin²y= e^2x, which is nonzero, so every point has a neighborhood in which f is one-to-one.

(3) This is where I'm really stuck. I assume we need to find a point that is mapped by two different ordered pairs but I don't know how to go about finding those points.

Any help on #3 or comments on the previous parts is appreciated. Thanks.

 

[HallsofIvy]

Think about $y_2= y_1+ 2\pi$.

 

[linearfish]

Oh, duh. I guess it didn't occur to me to hold x constant, so the points (0,pi) and (0,3*pi) both map to (-1,0), hence f is not one-to-one. Thanks.

Does the rest look okay?

 

[konthelion]

(1) and (2) looks fine to me.