Sorry but there is no standard way to do these.
Have a look at THIS SOLUTION.
$\displaystyle z^3=-3 i \overset{-}{z}$
Take absolute value
$\displaystyle \left|z|^3=\text{3$|$z$|$}\right.$
$\displaystyle |z|=0 \text{or} \sqrt{3}$
$\displaystyle z=0$ is a solution
To find the other solutions multiply the original equation by $\displaystyle z$
$\displaystyle z^4=-9i$
This can be solved to give four more solutions
Write z= a+ bi with a and b real. Then $\displaystyle z^3= (a+ bi)^3= a^3+ 3a^2bi+ 3a(bi)^2+ (bi)^3= a^3+ 3a^2bi- 3ab^2- b^3i= (a^3- 3ab^2)+ (3a^3b- b^3)i$. The right side of the equation is $\displaystyle -3\overline{z}= -3(a- bi)= -3a+ 3bi$. Equating real and imaginary parts, $\displaystyle a^3- 3ab^2= -3a$ and $\displaystyle 3a^2b- b^3= 3b$. One obvious solution is a= b= 0 so z= 0. Also if a= 0, the second equation becomes $\displaystyle -b^3= 3b$ and if b is not 0, $\displaystyle -b^2= 3$ which, since b is real, is impossible. If b= 0, the first equation becomes $\displaystyle a^3= -3a$ and if a is not 0, $\displaystyle a^2= -3$, again, impossible. That is, z= 0 is a solution and all other solutions must have both a and b non-zero.
If a and b are non-zero, we can divide $\displaystyle a^3- 3ab^2= -3a$ by a to get $\displaystyle a^2- 3b^2= -3$ and we can divide $\displaystyle 3a^2b- b^3= 3b$ by b to get $\displaystyle 3a^2- b^2= 3$. Adding those two equations, $\displaystyle 4a^3- 4b^3= 0$ so $\displaystyle a^3- b^3= (a- b)(a^2+ ab+ b^2)= 0$. If $\displaystyle a- b= 0$ then $\displaystyle a= b$ and we can write $\displaystyle 3a^2- b^2= 3$ as $\displaystyle 3a^2- b^2= 2a^2= 3$. $\displaystyle a^2= \frac{3}{2}$ so $\displaystyle a= \pm\sqrt{\frac{3}{2}}= \pm\frac{\sqrt{6}}{2}$. Two more solutions are $\displaystyle z= \frac{\sqrt{6}}{2}(1+ i)$ and $\displaystyle -\frac{\sqrt{6}}{2}(1+ i)$.
If $\displaystyle a\ne b$ then we must have $\displaystyle a^2+ ab+ b^2= 0$. By the quadratic formula, $\displaystyle a= \frac{-b\pm\sqrt{b^2- 4b^2}}{2}= \left(\frac{-1\pm\sqrt{-3}}{2}\right)b= \left(-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i\right)b$. But that is impossible because a and b are both real.
So the solutions to this equation are $\displaystyle z= 0$, $\displaystyle z= \frac{\sqrt{6}}{2}(1+ i)$, and $\displaystyle z= -\frac{\sqrt{6}}{2}(1+ i)$.
yes.
Let $\displaystyle z=r e^{i \theta }$
Take absolute value to show that $\displaystyle r^3=3r$ and so either $\displaystyle r=0$ or $\displaystyle r=\sqrt{3}$
$\displaystyle e^{3i \theta }= e^{i 3\pi /2} e^{-i \theta }=e^{i(3\pi /2-\theta )}$
and
$\displaystyle 3 \theta =3\pi /2-\theta +2 k \pi $
$\displaystyle \theta =\frac{3\pi }{8}+\frac{\pi }{2}k$
where $\displaystyle k=0,1,2,3$