Vikram's Circles

The two conditions can be written as a pair of (highly nonlinear) simultaneous equations for the modulus $r$ and argument $\theta$ of $z$ , since $|1-z^n| \; = \; \sqrt{1 + r^{2n} - 2r^n \cos n\theta}$

We know that $r < 1$ , but $r \approx 1$ , and that $\theta < \tfrac14\pi$ , with $\theta \approx \tfrac14\pi$ .

There are many methods for solving simultaneous equations numerically. A simple one is a multidimensional version of Newton-Raphson - see here . This method is not perfect, and so the NR method has to be shored up to make it more reliable.

I must admit that I am not hugely interested in the nitty-gritty of numerical calculation - I leave it to those who are - and so Mathematica's NSovle and FindInstance are my favourite methods!

The link I gave to Pi Han has the details. We are trying to solve the simultaneous equations $\begin{aligned} f_1(r,\theta) \; = \; (1+r)^2(1 - 2r^8\cos8\theta + r^{16}) - (1 + r^8)^2(1 - 2r\cos\theta + r^2) & = 0 \\ f_2(r,\theta) \; = \; (1+r)^2(1 - 2r^9\cos9\theta + r^{18}) - (1 + r^9)^2(1 - 2r\cos\theta + r^2) & = 0 \end{aligned}$ for $r \approx 0.93$ , $\theta \approx 0.73$ (these values represent a reasonable guess). I square everything in these equations to make $f_1,f_2$ differentiable. The two-dimensional Newton-Raphson iteration scheme makes us consider $\left(\begin{array}{c} r_{n+1} \\ \theta_{n+1}\end{array}\right) \; = \; \left(\begin{array}{c} r_{n} \\ \theta_{n}\end{array}\right) - J(r_n,\theta_n)^{-1}\left(\begin{array}{c} f_1(r_n,\theta_n) \\ f_2(r_n,\theta_n)\end{array} \right)$ where $J$ is the Jacobi matrix $J \; = \; \left(\begin{array}{cc} \frac{\partial f_1}{\partial r} & \frac{\partial f_1}{\partial \theta} \\ \frac{\partial f_2}{\partial r} & \frac{\partial f_2}{\partial \theta} \end{array} \right)$ While the 2D Newton-Raphson can be delicate, it works just fine in this case. If we choose $\binom{r_1}{\theta_1} \; = \; \binom{0.93}{0.73}$ then I get the values for $r$ and $\theta$ I mentioned in my proof, with $6$ -digit accuracy, after four rounds of iteration.