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!
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$$\omega^3=1\implies\omega\in\{1,-\frac{1}{2}+\frac{1}{2}i\sqrt{3},-\frac{1}{2}-\frac{1}{2}i\sqrt{3}\}$$
$$|z|=1$$ $\ $
$$\frac{z^{2(1+\omega)}-z^{2\omega}}{i}-e^{-\frac{\pi}{2\sqrt{3}}}=0\tag{1}$$
We see, equation (1) is an algebraic equation over $\mathbb{C}$ of more than one algebraically independent monomials ($z^{2(1+\omega)},z^{2\omega}$). The main theorem in [Ritt 1925] cannot help in these cases: We cannot read an elementary inverse function directly from the equation because we don't know how to rearrange the equation for $z$ by applying only finite numbers of elementary functions (operations) we can read from the equation. We have to guess $z$ therefore.
1) Solutions in terms of Elementary functions
1a) $\omega=1$
$$\frac{z^4-z^2}{i}-e^{-\frac{\pi}{2\sqrt{3}}}=0$$
We see, this equation is a $\mathbb{C}$-algebraic equation of $z$.
$$z\in\left\{\frac{1}{2}\sqrt{2+2\sqrt{1+4e^{-\frac{\pi}{2\sqrt{3}}}i}},-\frac{1}{2}\sqrt{2+2\sqrt{1+4e^{-\frac{\pi}{2\sqrt{3}}}i}},\frac{1}{2}\sqrt{2-2\sqrt{1+4e^{-\frac{\pi}{2\sqrt{3}}}i}},-\frac{1}{2}\sqrt{2-2\sqrt{1+4e^{-\frac{\pi}{2\sqrt{3}}}i}}\right\}$$
There is no $z$ with $|z|=1$.
1b) $z=\sqrt[6]{-1}$
Tyma Gaidash wrote in his answer, numerically, $z=\sqrt[6]{-1}$ is a solution.
$$z=\sqrt[6]{-1}\implies z\in\left\{-i,i,-\frac{1}{2}\sqrt{2+2i\sqrt{3}},\frac{1}{2}\sqrt{2+2i\sqrt{3}},-\frac{1}{2}\sqrt{2-2i\sqrt{3}},\frac{1}{2}\sqrt{2-2i\sqrt{3}}\right\}$$
Numerically, we find that $$z=\frac{1}{2}\sqrt{2+2i\sqrt{3}}$$ is a solution of the equation for $$\omega=-\frac{1}{2}+\frac{1}{2}i\sqrt{3}$$ with $|z|=1$.
2) Solutions in terms of nonelementary functions
$$\frac{z^{2(1+\omega)}-z^{2\omega}}{i}-e^{-\frac{\pi}{2\sqrt{3}}}=0\tag{1}$$
We see, equation (1) is a trinomial equation with complex exponents.
$$z^{2(1+\omega)}-z^{2\omega}-ie^{-\frac{\pi}{2\sqrt{3}}}=0$$ $$-\frac{1}{i}e^{\frac{\pi}{2\sqrt{3}}}z^{2\omega}+\frac{1}{i}e^{\frac{\pi}{2\sqrt{3}}}z^{2(1+\omega)}-1=0$$ $z\to e^{-\frac{1}{2}\frac{\pi}{2\sqrt{3}\omega}}(-xi)^\frac{1}{2\omega}$: $$x-e^{\frac{\pi}{\omega}i}e^{-\frac{\pi}{2\sqrt{3}\omega}}i^\frac{1}{\omega}x^{\frac{1}{\omega}+1}-1=0$$
Now the equation is in the form of equation 8.1 of [Belkic 2019]. Solutions in terms of Bell polynomials, Pochhammer symbols or confluent Fox-Wright Function $\ _1\Psi_1$ can be obtained therefore.
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 $-1$.