# 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!

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

$$Re(z)\geq Im(z)\iff x\ge y$$

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 $$1-2i$$.