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.
Boundary of the image of the unit disk under the exponential map.
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real-analysis
complex-analysis
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0Well, $$\begin{align}\bigl(y(t)\bigr)^2 &= \exp(2t)\sin^2\left(\sqrt{1-t^2}\right)\\ &= \exp(2t)\left(1-\cos^2\left(\sqrt{1-t^2}\right)\right)\\ &= \exp(2t)-\bigl(x(t)\bigr)^2,\end{align}$$ 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). – 2012-12-17