Equation for $x^2+y^2=1$ in exponential space?

Question

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.

Answer 1

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.