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The origin is to find the connected component of $e^z,|z|=1$, as $|z|=1$ is connected,and $e^z$ is continuous, $e^z$ should be a connect and the number is $1.$ I'm concerned with the image of $e^z$,let $u+iv$=$e^z=e^{x+iy}(x^2+y^2=1)$,so $u=e^x\cos y$ ,$v=e^x\sin y$

$u^2+v^2=e^{2x},\frac{v}{u}=\tan y$. It seems like a cardioid but I don't know how to draw it accurately.

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4  
Here it is. –  1015 Oct 12 '13 at 16:36

2 Answers 2

Parametrizing $z$ with $\cos t+i\sin t$, you have the equations $r=e^{\cos(t)}$, $\theta=\sin(t)$. Notice that even as $t$ spans through $[-\pi,\pi]$, $\theta$ will only span through $[-1,1]$.

For $t\in[-\pi/2,\pi/2]$, $t=\arcsin\theta$, so then the two equations can be used to eliminate $t$ and one segment of the curve is given by the polar equation $r=e^{\sqrt{1-\theta^2}}$ for $\theta\in[-1,1]$.

For other values of $t$ in $[-\pi,\pi]$, $t=\pi-\arcsin\theta$, and a similar simplification gives $r=e^{-\sqrt{1-\theta^2}}$ over the same $\theta\in[-1,1]$. The union of these two segments gives the moon shape in the WolfralAlpha link in the comments.

These two equations can be combined into a single implicit polar equation: $|\ln(r)|=\sqrt{1-\theta^2}$ for $\theta\in[-1,1]$, although the separate pieces $r=e^{\pm\sqrt{1-\theta^2}}$ are nicer because they are explicit polar equations.

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A graph from Mathematica yields enter image description here

It doesn't appear like any particular shape. You weren't too far off with your cardioid guess though.

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