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.
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$\begingroup$ 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). $\endgroup$– Cameron BuieDec 17, 2012 at 1:16
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$x^2 + y^2 = \exp(2t)$ 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 $x$.
answered Dec 17, 2012 at 3:25