Find all the roots of this complex equation

Question

Let $C$ be the set of complex numbers and $j$ the imaginary unit. Find all the roots(in $z$ $\in$ $C$) of the following equation:

$$ 2z^7 + 6z^4 = z^3e^{-j{\frac π7}} + 3e^{-j{\frac π7}} $$

Factorizing,

$$ 2z^4(z^3+3) = e^{-j{\frac π7}}(z^3+3) $$

1) If $$ z^3+3 = 0, z^3 = -3 = 3(cos\pi+jsin\pi) = 3e^{j\pi} $$

The solution then brings to, which I am not sure of. Anyone can explain this?

$$z_1 = \sqrt[3]{3}e^{j{\frac π3}}, z_2 = \sqrt[3]{3}e^{j\pi}, z_3 =\sqrt[3]{3}e^{j{\frac {5π}3}} $$

if $$ z^3 + 3 \ne 0$$ then $$2z_4 =e^{-j{\frac π7}}, z^4 =\frac 12 e^{-j{\frac π7}} = \frac 12 e^{j{\frac {13\pi}7}} $$

$$ z_4 = \sqrt[4]{\frac12}e^{j{\frac {13\pi}{28}}}, z_5 = \sqrt[4]{\frac12}e^{j{\frac {27\pi}{28}}}, z_6 = \sqrt[4]{\frac12}e^{j{\frac {41\pi}{28}}}, z_7 = \sqrt[4]{\frac12}e^{j{\frac {55\pi}{28}}}$$

Answer 1

If $z^n = re^{j\theta}$ with $r >0$, then $$z= (re^{j\theta})^{1/n}\omega^k, k =0,1,2,\cdots, n-1$$ where $\omega = \exp{\frac{j2\pi}{n}}$, since if you take $z= (re^{j\theta})^{1/n}y$, the problem reduces to solve $y^n = 1$