Equation for $x^2+y^2=1$ in exponential space? Consider the unit circle $x^2+y^2=1$ in exponential space. Is there an equation for the set of points under the mapping?
Here's what I tried:
I took each coordinate pair on the unit circle and exponentiated it. 
For example, $(x,y)\mapsto(e^x,e^y).$
This is what I have so far:

I also tried to come up with an equation for this set of points but have not been able to so far.

The equation for the set of points under the prescribed mapping is $\ln(x)^2+\ln(y)^2=1.$
Start with $(x,y) \mapsto (e^x,e^y).$ Define $u=\ln(x)$ and $v=\ln(y).$ In the pre-image space, we simply have $x^2+y^2=1.$ In the image space (after the prescribed mapping) we get $u^2+v^2=1.$ Substituting back to get an explicit equation in terms of $x$ and $y$ we get the above equation $\ln(x)^2+\ln(y)^2=1.$
Notice that in a $u-v$ coordinate system, we actually do have the equation for a circle, but using an $x-y$ coordinate system for the image space, we get a nonlinear equation in the coordinates $x-y.$ So if you ever have to deal with $\ln^2(x)+\ln^2(y)=1,$ just remember that you can always simplify it down to the equation of a circle using a change of coordinates which acts to linearize the equation from the image space to the pre-image space.
