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Let $z_1$ and $z_2$ be 2 elements of $C _∞$ . Determine the set of point $S$ in $C_∞$ such that the corresponding set $S'$ on the sphere is a circle that is equidistant from $z_1'$ and $z_2'$ (the point on the sphere corresponding to $z_1$ and $z_2$)

I'm not sure if I understand the question correctly, but here is what I got

$f(S)=S'$, $f(z_1)=z_1'$ and $f(z_2)=z_2'$.

Now to the confused part, $S'$ on the sphere is a circle that is equidistant from $z_1'$ and $z_2'$ meaning $|S-z_1'|=|S-z_2'|$. So if we have 2 circles with same radius and center $z_1'$ and $z_2'$ then $S'$ is the intersect points of these 2 circles? Then what about $S'$ is a circle? Can I imagine the sphere is like the earth then $S'$ is like the equator and $z_1'$ is the north pole , called $N=(0,0,1)$ and $z_2'$ is the south pole $SO=(0,0,-1)$?

Mister Benjamin Dover
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Diane Vanderwaif
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1 Answers1

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On pages 20 and 50 of Ahlfors, you will see the distance in the extended complex plane is defined as $$ d(z_1,z_2) = \frac{2\lvert z_1-z_2\rvert}{\sqrt{(1+\lvert z_1\rvert^2)(1+\lvert z_2\rvert^2)}}. $$ A point equidistance from $z_1$ and $z_2$ will satisfy $$ \frac{\lvert z -z_1\rvert}{\lvert z -z_2\rvert} = \frac{\sqrt{1+\lvert z_1\rvert^2}}{\sqrt{1+\lvert z_2\rvert^2}}. $$ Consider two cases $\lvert z_1\rvert = \lvert z_2\rvert$ and $z_2=\infty$. When they are equal, we simply have $$ \lvert z -z_1\rvert=\lvert z -z_2\rvert. $$ Additionally, on page 20 Ahlfors, we have that when $z_2=\infty$ $$ d(z_1,\infty)=\frac{2}{\sqrt{1+\lvert z_1\rvert^2}} $$ so the points satisfying this are $$ \lvert z -z_1\rvert = \sqrt{1+\lvert z_1\rvert^2} $$

dustin
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