Ok here's the problem:
F: R^2 -> R^2
(t is theta)
F(r,t) = (r*cos(t), r*sin(t))
What is F^-1(0,0)?
I know the answer is {(0,0) | -inf < t < inf }
but I'm not exactly sure how to get to that point.
lets start with this
$\displaystyle F(r,\theta)=(r\cos(\theta),r\sin(\theta))$
consider $\displaystyle F(r,\theta)=(0,0)$
so we would need to know when
$\displaystyle r\cos(\theta)=0 \mbox{ and } r\sin(\theta)=0$
since cosine and sine are never zero at the same value of theta r must be zero. If that is the case then the equation
$\displaystyle 0\cos(\theta)=0 \mbox{ and } 0\sin(\theta)=0$
is true for all values of theta so the inverse immage of (0,0) is
$\displaystyle [{(r,\theta)|r=0, -\infty < \theta < \infty}]$