# Boundary of the image of the unit disk under the exponential map.

Anyone know the Cartesian coordinate equation for the top half of the boundary of the image of the unit disk under the exponential map in $\mathbb{C}$? Finding parametric equations in Cartesian coordinates was easy enough: $$x(t)=\exp(t)\cos(\sqrt{1 - t^2})\\ y(t)=\exp(t)\sin(\sqrt{1 - t^2})\\ -1\leq t\leq 1$$ I was also able to describe the curve in polar coordinates: $$r(\theta)=\exp(\pm\sqrt{1-{\theta}^2})\\ -1\leq \theta\leq 1$$ I haven't had any luck finding an equation for $y$ as a function of $x$ for the top half of this curve, however.

• Well, \begin{align}\bigl(y(t)\bigr)^2 &= \exp(2t)\sin^2\left(\sqrt{1-t^2}\right)\\ &= \exp(2t)\left(1-\cos^2\left(\sqrt{1-t^2}\right)\right)\\ &= \exp(2t)-\bigl(x(t)\bigr)^2,\end{align} so since we're dealing with $y(t)\geq 0$, then $y(t)=\sqrt{\exp(2t)-\bigl(x(t)\bigr)^2}$. Unfortunately, this still depends on $t$ (and there may be no way out of that). Dec 17, 2012 at 1:16

$x^2 + y^2 = \exp(2t)$ so $t = \ln \sqrt{x^2+y^2}$. Also $\tan \sqrt{1-t^2} = y/x$ with $0 \le \sqrt{1-t^2} \le 1$ so $\arctan(y/x) = \sqrt{1-t^2}$. Thus you have the implicit equation

$$\left(\ln \sqrt{x^2+y^2}\right)^2 + \left(\arctan(y/x)\right)^2 = 1$$

You won't get a closed-form solution for $y$ as a function of $x$.