I need a closed form for the zeros of $$f(z):=\frac{z^{2(1+\omega)}-z^{2\omega}}{i}-e^{-\frac{\pi}{2\sqrt{3}}} $$ where $z$ lies on the unit circle $|z|=1$ and $\omega$ is a cube root of unity.
We know that the cube roots of unity satisfy $$1+\omega+\omega^2=0\ \ \ \text{and}\ \ \ \omega^3=1$$
Now we have
$$\frac{z^{2(1+\omega)}-z^{2\omega}}{i}-e^{-\frac{\pi}{2\sqrt{3}}}=0$$
So $$e^{\frac{\pi}{2\sqrt{3}}}z^{2\omega}\left(\frac{z^2-1}{i}\right)-1=0$$
Any help would be highly appreciated. Thanks!
Best Answer
Numerically, $z=\sqrt[6]{-1}$ is the only $|z|=1$ solution to $$z^{1+ i\sqrt 3}-z^{-1+ i\sqrt 3}=ie^{-\frac\pi{2\sqrt 3}}$$
Now substituting $z=\pm\sqrt w$ and use Lagrange reversion for the other equation:
$$\begin{align}z^{1- i\sqrt 3}-z^{-1- i\sqrt 3}=ie^{-\frac\pi{2\sqrt 3}}\iff z^2=i e^{-\frac\pi{2\sqrt3}}z^{2\sqrt[3]{-1}}+1\iff w= (\pm 1)^{2\sqrt[3]{-1}}i e^{-\frac\pi{2\sqrt3}}w^\sqrt[3]{-1}+1\implies z=\pm 1\pm\frac12\sum_{n=1}^\infty\frac{\left(i(\pm 1)^{2\sqrt[3]{-1}}e^{-\frac\pi{2\sqrt3}}\right)^n}{n!}\left.\frac{d^{n-1}}{dw^{n-1}}w^{\sqrt[3]{-1}n-\frac12}\right|_1\end{align}$$
to finally use the Fox Wright function, a special case of Fox H:
$$\bbox[5pt,border: 1px solid blue]{\begin{align}z=\pm1\pm\frac12\sum_{n=1}^\infty\frac{\left(\pm i e^{-\frac{\{1,7\}\pi}{2\sqrt 3}}\right)^n}{n!}\frac{\Gamma\left(\frac12+\sqrt[3]{-1}n\right)}{\Gamma\left(\frac32+(-1)^\frac23n\right)}=\pm1\pm\frac12\,_1\Psi_1\left(^{\left(\frac12,\sqrt[3]{-1}\right)}_{\left(\frac32,(-1)^\frac23\right)};\pm i e^{-\frac{\{1,7\}\pi}{2\sqrt 3}} \right)\end{align}}$$
where one solution takes $+,1$ and the other takes $-,7$. Both roots match Wolfram Alpha’s. This calculates $z+1$ more precisely for $z$ being the root taking $-,7$ as it’s real part is almost $-1$.