Determine the number of point $z \in \mathbb{C}$ such that $(2z+i\overline{z})^3=27i$ I've just learned complex numbers in Mathematical Analysis 1, and I'm stuck in the following problem: I would like to determine the number of point  $z \in \mathbb{C}$ such that $(2z+i\overline{z})^3=27i$, and solve the following system of equations:
$\begin{cases}\begin{matrix} (2z+i\overline{z})^3=27i \\ Re(z)\geq Im(z) \end{matrix}\end{cases}$. 
Can someone help me explaining in detail the steps? Thank you very much! 
 A: HINT
We have that
$$(2z+i\overline{z})^3=27i \iff 2z+i\bar z=3\sqrt[3]i$$
and for each solution for $w=\sqrt[3]i\,$ that is


*

*$w_1=\frac{\sqrt 3}2+\frac12 i$

*$w_2=-\frac{\sqrt 3}2+\frac12 i$

*$w_3=- i$
we can determine $z=x+iy\,$ and the select the solutions which satisfy
$$Re(z)\geq Im(z)\iff x\ge y$$
A: The LHS is under a cube, so we need to take the cube root of $27i$. In polar form this is $27e^{i\pi/2}$, so the principal cube root is $3e^{i\pi/6}$ and the other two roots are obtained by multiplying by $e^{2i\pi/3}$ (one-third of a revolution).
This yields, upon converting back to Cartesian form:
$$3e^{i\pi/6}=\frac{3\sqrt3}2+i\frac32$$
$$3e^{i5\pi/6}=-\frac{3\sqrt3}2+i\frac32$$
$$3e^{i3\pi/2}=-3i$$
We are now told that these are equal to $2z+i\overline z$. Let $z=x+iy$, then
$$2z+i\overline z=2(x+iy)+i(x-iy)=(2x+y)+(2y+x)i$$
For each cube root of $27i$ found above, this is a linear system equating the real and imaginary parts. The solutions are
$$\sqrt3-\frac12+\left(1-\frac{\sqrt3}2\right)i$$
$$-\sqrt3-\frac12+\left(1+\frac{\sqrt3}2\right)i$$
$$1-2i$$
The last condition on the real part being greater than the imaginary part gives the solutions as $\sqrt3-\frac12+\left(1-\frac{\sqrt3}2\right)i$ and $1-2i$.
