- #1
linearfish
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Homework Statement
Let f = (f1,f2) be the mapping of R2 into R2 given by f1=excos(y), f2=exsin(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 ex 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 = e2xcos2y + e2xsin2y= e2x, 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.