# Radius of the spherical image of a circle

This is question 5 on page 20 of the book Complex Analysis by Lars Ahlfors. I have no idea how to answer that problem:

Find the radius of the spherical image of the circle in the plane whose center is $a$ and radius is $R$.

Here spherical image means: the image of a subset of complex numbers under the identification of the complex plane with the sphere $\Bbb S^2$ (the Riemann sphere) by stereographic projection:

Thanks.

• You need to post more details about this problem. Very few people will have this book. In order to discuss an image, we need to know what the map is. Sep 3 '12 at 1:03
• @DrKW the map is the stereographic projection,also one should know the metric induced by the Riemann sphere to understand the problem.regards.
– GjR
Sep 3 '12 at 2:08
• I used to have Ahlfors’s book, but I forget whether the plane is placed so as to pass through the center of the sphere or so as to be tangent to the sphere at the lowest point. This does affect the answer, so I have to agree with @Kris. Apr 18 '13 at 1:33
• @Lubin Thanks for point that out. The plane is placed as in the image.
– leo
Apr 18 '13 at 2:02
• Below I read an answer for balls in $\Bbb C$ (which should be $\overline{\Bbb C}$ I would say) of the form $\{|z-a|<R\}$. What about the other balls $\{|z-a|>R\}$ in $\overline{\Bbb C}$?
– Joe
Jul 27 '20 at 17:01

As you can see from the picture, the geometry is symmetric about the vertical axis, in other words the answer depends only on $R$ and $A=|a|$. So if we take our circle to be centered on the $x$-axis, as @Hagen has suggested, we only need to look at the intersection of the picture with the $(x,z)$-plane. The map from the $x$-axis to the circle $x^2+z^2=1$ is: $$\xi\mapsto \left(\frac{2\xi}{\xi^2+1},\frac{\xi^2-1}{\xi^2+1}\right)\,.$$ Now you have to plug in $A+R$ and $A-R$ for $\xi$, find the two points the formula gives you, and the distance between them is the diameter of the circle you want, again making use of @Hagen’s suggestion. Looks like very messy algebra, and I do wonder whether there’s a slicker way to do it.

I suppose I should boast that I learned this stuff out of Ahlfors’s book, with Ahlfors himself standing in front of the class. He was a superb teacher.

• The algebra is not that messy compared with what I was doing. This is fine, the only problem to me is, why is it clear that diametrically opposite points of the circle in the complex plane corresponds to diametrically opposite points in the circle on the $xz$ plane?
– leo
Apr 18 '13 at 4:14
• I’m not sure I’m even convincing myself with this argument, but the whole situation is symmetric with respect to the real (that is, the $x$-) axis, whose image under stereography is the great circle gotten by intersecting our sphere with the $(x,z)$-plane. This great circle clearly bisects the circle lying on the sphere, ’cause again this circle is mapped to itself by the involution $(x,y,z)\mapsto(x,-y,z)$. Apr 19 '13 at 5:08
• @leo: yup's argument also looks convincing once you know that the stereographic projection maps circles to circles: the circle $C$ of radius $R$ around $A$ has two tangent circles: the circles around the origin through $A+R$ and $A-R$. These two tangent circles are mapped to circles of latitude of the sphere which are tangent to the image of $C$. Circles of latitude have horizontal tangents, so the image of $C$ has parallel tangents in the images of $A+R$ and $A-R$, so these are diametrically opposite. Instead of exploiting symmetry at the $(x,z)$-plane, this exploits rotational symmetry. Apr 19 '13 at 9:11
• @Martin I do know that projections of circles are circles or straight lines. The other way arround, circles in the complex plane are mapped to circles in the sphere, and straight lines are mapped to circles through the north pole, so what you say makes sense. Thanks for your explanation.
– leo
Apr 19 '13 at 19:32
• Okay now I'm confused. I can see that the image of a circle centered at real axis lies in the $xz$ plane because the $y$ coordinate of its points is $0$. I've seen the proof that the image of circles in the plane are circles in the sphere. But if imagine the intersection of the unit sphere and the $xz$ plane, what I see is just a circle of radius $1$. So, since the image of a circle is a circle, the image of this circle centered at the real axis must be the whole unit circle. Please helpme.
– leo
Apr 19 '13 at 20:20

Ahlfors gives a slick derivation of the chordal distance formula $$d(z,z') = \frac{2|z-z'|}{\sqrt{(1+|z|^2)(1+|z'|^2)}}$$ on page 20 of his book. Here $z,z' \in \mathbb C$ and the chordal distance is by definition the distance of $z$ to $z'$ after applying stereographic projection to them.

Taking $z= A+R$ and $z'=A-R$ as suggested by @lubin this simplifies to $$d(A+R,A-R) = \frac{2R}{\sqrt{(1+(A+R)^2)(1+(A-R)^2)}} = \frac{2R}{\sqrt{1+2A^2+2R^2+(A^2-R^2)^2}}$$ and dividing this by $2$ yields the formula you're looking for.

Since you ask in a comment how to see that the diametrically opposite points $A+R$ and $A-R$ are mapped to diametrically opposite points: observe that the circles around the origin through $A\pm R$ are tangent to the circle $C$ of radius $R$ around $A$. The stereographic projection maps them to circles of latitude tangent to the image of the circle $C$.

The algebra is not all that messy:

Let $x = (x_1,x_2,x_3)$ and $x'=(x_1',x_2',x_3')$ be the points on $S^3$ corresponding under stereographic projection to $z$ and $z'$, respectively. Then $\lVert x\rVert^2 = 1=\lVert x'\rVert^2$ and \begin{align*} \lVert x - x'\rVert^2 &= (x_1 - x_1')^2 + (x_2-x_2')^2 + (x_3-x_3')^2 \cr &= 2 - 2(x_1 x_1' + x_2 x_2' + x_3 x_3'). \end{align*} After reminding the reader of $\DeclareMathOperator{\Re}{Re}\DeclareMathOperator{\Im}{Im}$ $$(x_1,x_2,x_3) = \frac{1}{1+|z|^2}\left(2\Re z, 2\Im z, |z|^2-1\right) = \frac{1}{1+|z|^2}\left(z+\bar{z}, z-\bar{z}, |z|^2-1\right),$$ Ahlfors computes \begin{align*} x_1 x_1' + x_2 x_2' + x_3 x_3' &= \frac{(z+\bar{z})(z'+\bar{z'}) + (z-\bar{z})(z'-\bar{z'}) + (|z|^2-1)(|z'|^2-1)}{(1+|z|^2)(1+|z'|^2)} \cr &= \frac{(1+|z|^2)(1+|z'|^2) - 2|z-z'|^2}{(1+|z|^2)(1+|z'|^2)} \end{align*} and obtains $$\lVert x - x'\rVert = \frac{2|z-z'|}{\sqrt{(1+|z|^2)(1+|z'|^2)}}.$$

• There is a minor error in the above displayed formulae. Twice the imaginary part of a complex number is $i$ multiplied by the difference between a complex number and its conjugate. Consequently, the second term in the {\it{numerator}} after the comment "Ahlfors computes.." which is $+(z-{\bar{z}})(z'-{\bar{z'}})''$ should have a {\it{negative}} sign in front of it, and read $-+(z-{\bar{z}})(z'-{\bar{z'}})''$. This is verified by the computation in Ahlfors's book (page 20). Apr 19 '17 at 14:06
• I believe the numerator of $d(A+R,A−R)$ should be $4R$ not $2R$, because $|z-z'| = 2R$. Otherwise +1 for the most elegant approach in my opinion! May 25 '18 at 10:00

I think I managed to write it properly.

Let's denote the stereographic projection from $$\Bbb C$$ to the sphere by $$\varphi$$. Let's denote $$C(a,R)$$ the circle with center $$a$$ and radius $$R$$.

Assume first that $$a\in[0,\infty[$$. Then since $$\varphi$$ maps circles into circles, the diameter of the circle image is $$D = \sup\{d(\varphi(z),\varphi(a+R)):z\in C(a,R)\}.$$

The approach is to find a bound for $$\{d(\varphi(z),\varphi(a+R)):z\in C(a,R)\}$$

and then see that the bound is reached at a point in the circle so it has to be the sup.

So, observe that

\begin{align} d(\varphi(z),\varphi(a+R)) &= \frac{2\lvert z - (a+R)\rvert}{\sqrt{(1+\lvert z\rvert^2)(1+\lvert a + R\rvert^2)}} \\ &\leq \frac{2(\lvert z-a\rvert + R)}{\sqrt{(1+\lvert z\rvert^2)(1+\lvert a + R\rvert^2)}} \\ &= \frac{4R}{\sqrt{(1+\lvert z\rvert^2)(1+\lvert a + R\rvert^2)}} \tag{1} \end{align}

In the other hand, if $$z = x + iy$$ is in $$C(a,R)$$ then $$R^2 = (x-a)^2 + y^2,$$

so

\begin{align} \lvert z \rvert^2 &= x^2 + R^2 - (x-a)^2 \\ &= x^2 + R^2 - x^2 + 2xa - a^2 \\ &= R^2 + 2xa - a^2\tag{2} \end{align}

but since $$z$$ is in the circle $$\Re z = x \in [a-R,a+R]$$. So by (2) \begin{align} \lvert z \rvert^2 &= R^2 + 2xa - a^2 \\ &\geq R^2 + 2(a-R)a - a^2 \\ &= (a-R)^2. \end{align}

So

\begin{align} 1 + \lvert z\rvert^2 &\geq 1 + (a-R)^2 \\ \frac{1}{\sqrt{1 + \lvert z\rvert^2}} &\leq \frac{1}{\sqrt{1 + (a-R)^2}}. \end{align}

By (1)

$$d(z,a+R) \leq \frac{4R}{\sqrt{(1+\lvert a - R\rvert^2)(1+\lvert a+R\rvert^2)}}$$

and the bound is reached at $$z=a-R$$, so the radius is $$\frac{2R}{\sqrt{(1+\lvert a - R\rvert^2)(1+\lvert a+R\rvert^2)}}.$$

Now, notice that the formula $$d(\varphi(z),\varphi(z'))=\frac{2|z-z'|}{\sqrt{(1+|z|^2)(1+|z'|^2)}}$$ is invariant under rotations.

So if $$a\in\Bbb C \setminus [0,\infty[$$, in particular it is not $$0$$, so rotate by $$\frac{\bar a}{\lvert a \rvert}$$ and we are done.

One Approach: We want $F:\mathbb R^2 \to \mathbb S^2$ where $\mathbb S^2 = x^2+y^2+z^2=2$.

You can calculate the image of $(x,y)$.While trying to solve,I got something like: $$F(x,y)=\left(\frac{2x}{x^2+y^2+1},\frac{2y}{x^2+y^2+1},\frac{x^2+y^2-1}{x^2+y^2+1}\right)$$

Now,As far as I know,circles are sent to circles under this projection,so,get any 3 points of one circle and find their images.As any three points completely describe the cirle,we are done.

P.S:I am not sure about the formula,so please check it from the book you are using.

• By symmetry, consider a circle with center $a$ on the nonnegative real axis. Then the images of $a\pm R$ will be one diameter apart. Sep 8 '12 at 12:00
• Dear @HagenvonEitzen would you like to elaborate it in an answer? I've solved this but my solution involves a lot of computations.
– leo
Apr 17 '13 at 22:28
• The formula with $F$ is fine, but the equation for elements of $\Bbb S^2$ must be: $$x^2+y^2+z^2=1$$
– leo
Apr 18 '13 at 0:25