Complex numbers on circle of unit radius

Question

Given three points in the complex plane (i.e. numbers $z_1,z_2,z_3\in\mathbb C$), they define a unique circle (unless they are collinear). When does that circle have radius one?

I know how to compute that “the hard way”, i.e. by separating real and imaginary part. From there I could either construct perpendicular bisectors and intersect these, or I could solve $ax_k+by_k+c=x_k^2+y_k^2$ for $k\in\{1,2,3\}$ and then deduce the radius from the $a,b,c$ I found.

But I guess there might be some more elegant way to express this condition using vocabulary more suited for complex numbers. Separating numbers into real and imaginary part for all numbers should not be needed, even though conjugation might still be needed at some point.

As a motivating example: we know that four points are cocircular iff they satisfy

$$\begin{vmatrix} x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ x_3^2+y_3^2 & x_3 & y_3 & 1 \\ x_4^2+y_4^2 & x_4 & y_4 & 1 \end{vmatrix}=0$$

but with $z_k=x_k+iy_k$ you can also check the condition

$$\frac{(z_1-z_3)(z_2-z_4)}{(z_1-z_4)(z_2-z_3)}\in\mathbb R$$

which is a lot easier to write and compute. I'm looking for some similar simplification for the case of unit radius.

Answer 1

You could compute $r = A/2\sin\alpha$ which would be a bit easier, as $A = |c-b|$ and $$\sin\alpha = \left|\Im\left(\frac{(b-a)|c-a|}{(c-a)|b-a|}\right)\right|$$

But well, I'm aware this is not very elegant either...

Answer 2

The comment by Tom gave me an idea. This is no complete answer yet, so please expand if you can think of ways to proceed. Let's start with what he wrote:

$$\exists z\in\mathbb C:\quad \lvert z-z_1\rvert=\lvert z-z_2\rvert=\lvert z-z_3\rvert=1$$

Let's square things, and express absolute values using conjugation.

$$\exists z\in\mathbb C\,\forall k\in\{1,2,3\}:\quad 1=(z-z_k)(\bar z-\bar z_k)=z\bar z-z\bar z_k-\bar zz_k+z_k\bar z_k$$

Now we can treat $a=-\bar z$ and $b=-z$ as two distinct variables and obtain

$$ab + az_k + b\bar z_k + \lvert z_k\rvert^2 - 1 = 0$$

Three conditions for two variables, which is still too much. So let's treat $c=ab$ as a third variable.

$$az_k + b\bar z_k + c = 1 - \lvert z_k\rvert^2$$

So now we have a $3\times 3$ system of equations, and once we have solved that, we can check whether $c=ab$ holds. I guess that if it does, then $a=\bar b$ will hold as well. To verify that guess: For reasons of symmetry $c$ is real. Therefore $az_k$ and $b\bar z_k$ must have opposite imaginary parts. Furthermore, $az_k\cdot b\bar z_k=c\lvert z_k\rvert^2\in\mathbb R$ so they must have opposite argument as well. If two numbers have opposite imaginary part and opposite argument, then they are conjugate to one another.

Unfortunately solving a system of three linear equations is still harder than what I'd have hoped for. One big win is that this approach is very symmetric, and can be done symbolically as well. Unfortunately, the result still looks evil:

$$ z_1^2\bar z_1^2z_2\bar z_2 - z_1\bar z_1^2z_2^2\bar z_2 - z_1^2\bar z_1z_2\bar z_2^2 + z_1\bar z_1z_2^2\bar z_2^2 - z_1^2\bar z_1^2\bar z_2z_3 + \bar z_1^2z_2^2\bar z_2z_3 + z_1^2\bar z_1\bar z_2^2z_3 - \bar z_1z_2^2\bar z_2^2z_3 + z_1\bar z_1^2\bar z_2z_3^2 - \bar z_1^2z_2\bar z_2z_3^2 - z_1\bar z_1\bar z_2^2z_3^2 + \bar z_1z_2\bar z_2^2z_3^2 - z_1^2\bar z_1^2z_2\bar z_3 + z_1\bar z_1^2z_2^2\bar z_3 + z_1^2z_2\bar z_2^2\bar z_3 - z_1z_2^2\bar z_2^2\bar z_3 + z_1^2\bar z_1^2z_3\bar z_3 - \bar z_1^2z_2^2z_3\bar z_3 - z_1^2\bar z_2^2z_3\bar z_3 + z_2^2\bar z_2^2z_3\bar z_3 - z_1\bar z_1^2z_3^2\bar z_3 + \bar z_1^2z_2z_3^2\bar z_3 + z_1\bar z_2^2z_3^2\bar z_3 - z_2\bar z_2^2z_3^2\bar z_3 + z_1^2\bar z_1z_2\bar z_3^2 - z_1\bar z_1z_2^2\bar z_3^2 - z_1^2z_2\bar z_2\bar z_3^2 + z_1z_2^2\bar z_2\bar z_3^2 - z_1^2\bar z_1z_3\bar z_3^2 + \bar z_1z_2^2z_3\bar z_3^2 + z_1^2\bar z_2z_3\bar z_3^2 - z_2^2\bar z_2z_3\bar z_3^2 + z_1\bar z_1z_3^2\bar z_3^2 - \bar z_1z_2z_3^2\bar z_3^2 - z_1\bar z_2z_3^2\bar z_3^2 + z_2\bar z_2z_3^2\bar z_3^2 + \bar z_1^2z_2^2 - 2z_1\bar z_1z_2\bar z_2 + z_1^2\bar z_2^2 - 2\bar z_1^2z_2z_3 + 2z_1\bar z_1\bar z_2z_3 + 2\bar z_1z_2\bar z_2z_3 - 2z_1\bar z_2^2z_3 + \bar z_1^2z_3^2 - 2\bar z_1\bar z_2z_3^2 + \bar z_2^2z_3^2 + 2z_1\bar z_1z_2\bar z_3 - 2\bar z_1z_2^2\bar z_3 - 2z_1^2\bar z_2\bar z_3 + 2z_1z_2\bar z_2\bar z_3 - 2z_1\bar z_1z_3\bar z_3 + 2\bar z_1z_2z_3\bar z_3 + 2z_1\bar z_2z_3\bar z_3 - 2z_2\bar z_2z_3\bar z_3 + z_1^2\bar z_3^2 - 2z_1z_2\bar z_3^2 + z_2^2\bar z_3^2=0 $$

Still, there are some applications where this formulation is better than the naive approach of splitting real and imaginary part. Namely when you are working in some subfield of $\mathbb C$ where the imaginary part is not in that field and therefore computing it comes at a high conversion cost, e.g. in a cyclotomic field. The above can stay in that field the whole time, without conversions.

Answer 3

It seems the following.

Let the triangle with vertices $z_1, z_2, z_3$ has sides $a, b, c$, area $S$ and radius $R$ of the circumcircle. Then $R=\frac {abc}{4S}$. The formula for oriented area

$$2S=\left|\begin{array}{} x_1-x_3 & x_2-x_3\\ y_1-y_3 & y_2-y_3\\ \end{array}\right|, $$

yields $4S=2\Im(\bar z_1z_2+\bar z_2z_3+\bar z_3z_1)$. So $R=1$ iff $$|z_1-z_2||z_2-z_3||z_3-z_1|=2|\Im(\bar z_1z_2+\bar z_2z_3+\bar z_3z_1)|.$$