And it seem that,
\(a^{\frac{1}{m}}+b^{\frac{1}{n}}=c^\frac{1}{k}\)
is just,
\(cos(\theta_m)+cos(\theta_n)=cos(\theta_k)\), where the radius of the circle has been normalized with a factor \(\cfrac{\pi}{2}\).
\(r\theta_n=arc_n\) , \(r=\cfrac{2}{\pi}\)
\(\theta_n=\cfrac{1}{n}\cfrac{\pi}{2}\)
\(\theta_m=\cfrac{1}{m}\cfrac{\pi}{2}\)
\(\theta_k=\cfrac{1}{k}\cfrac{\pi}{2}\)
\(cos(\cfrac{\pi}{2m})+cos(\cfrac{\pi}{2n})=cos(\cfrac{\pi}{2k})\)
What happened to \(a^{m}, b^{n}, c^{k}\)?
$${ln(y)nln(a)}=\frac{\pi}{2}ln(a)$$
Consider, $$y=e^{i\theta}$$
$$ln(y)=i\cfrac{1}{n}\cfrac{\pi}{2}$$
$$\cfrac{2ln(y)ln(a)}{\pi}=i\frac{1}{n}ln(a)$$
$$ln(y)nln(a)=i\frac{\pi}{2}ln(a)$$
$$ln(y)=ln(a)(\frac{i\pi}{2}-n)$$
$$e^{i\theta}=a^{\cfrac{i\pi}{2}-n}$$
$$a^n=e^{-i\theta} a^{\cfrac{i\pi}{2}}$$
And so,
$$a^m=e^{-i\theta_m} a^{\cfrac{i\pi}{2}}$$
$$b^n=e^{-i\theta_n} a^{\cfrac{i\pi}{2}}$$
$$c^k=e^{-i\theta_k} a^{\cfrac{i\pi}{2}}$$
With \(\theta_n=\cfrac{\pi}{2n}\), \(\theta_m=\cfrac{\pi}{2m}\), \(\theta_k=\cfrac{\pi}{2k}\)
$$a^m+b^n=e^{-i\cfrac{\pi}{2n}}. a^{i\cfrac{\pi}{2}}+e^{-i\cfrac{\pi}{2m}} .b^{i\cfrac{\pi}{2}}=e^{-i\cfrac{\pi}{2k}}. c^{i\cfrac{\pi}{2}}=c^k$$
And they just keep turning.
