Determine all complex numbers which satisfy conditions - $|z|=2$ $\space$ and $\space$ Im$(z^6)=8$ Im$(z^3)$

It is shorter to solve with the exponential form of $z$: since its modulus is $2$, we can write $\:z=2\,\mathrm e^{i\theta}$. and the equation on the imaginary parts becomes $\DeclareMathOperator{\im}{Im}$ $$\im(z^6)=\im\bigl(64\mathrm e^{6i\theta} \bigr)=64\sin 6\theta,\qquad \im(8z^3)=\im\bigl(64\mathrm e^{3i\theta}\bigr)=64\sin 3\theta$$ whence this simple standard trigonometric equation $\;\sin 6\theta=\sin 3\theta$. Its solutions are $$ \begin{cases} 6\theta\equiv 3\theta \iff 3\theta\equiv 0\mod 2\pi\iff\theta\equiv 0\mod\frac{2\pi}3,\\ 6\theta\equiv \pi-3\theta \iff 9\theta\equiv \pi \mod 2\pi \iff \theta\equiv \frac\pi 9 \mod\frac{2\pi}9 . \end{cases} $$ A short form of the solutions in $\theta$ would be $$\theta\in\Bigl\{\frac{k\pi}9\,\Bigm|\, k=0, \pm 1,\pm3,\pm 5,\pm 7, 9 \Bigr\}. $$

Without loss of generality, we can reduce the equations to $$|z|=1 ,\text{Im}(z^6)=\text{Im}(z^3)$$

From this, we can say that when $z=\omega_i$ (where $\omega_i$ are the cube roots of unity), the equations will definitely be true.

After that, use the polynomial expansions for $z^6 $ and $z^3$ considering $z=x+i y$ which is effectively solving $$6xy^{5}\ -20x^{3}y^{3}+6x^{5}y=\left(3yx^{2}-y^{3}\right)$$ upon condition that $$x^2+y^2=1$$ which is a unit circle.

You can access the following graph here

The intersections of the black graph with the red circle and the blue points with coordinates labelled are the solutions required.