hint
If $$z=re^{i\theta}$$ then
$$z^n=r^n(\cos(n\theta)+i\sin(n\theta))$$
the real part gives
$$r^{2018}\cos(2018\theta)=2018^{2018}$$
and the imaginary
$$r^{2018}\sin(2018\theta)=1$$
Don't forget about the "$= 0$" part of the equations.
The formula $(ax^2+ay^2+2(a'x+b'y)+c,0)$ is not an equation;
the equivalent equation to $a\lvert z\rvert^2+\bar{b}z+b\bar{z}+c=0$ is
$$(ax^2+ay^2+2(a'x+b'y)+c,0) = (0,0),$$ or more simply
$$ax^2+ay^2+2(a'x+b'y)+c = 0.$$
For the equation $\lvert c\rvert z-c\lvert z\rvert=0,$
in your $\mathbb R^2$ notation you get the equation
$$ \left(\sqrt{q^2+r^2}x-\sqrt{x^2+y^2}q,\sqrt{q^2+r^2}y-\sqrt{x^2+y^2}r\right) = (0,0).$$
That is equivalent to the two simultaneous equations
\begin{align}
x\sqrt{q^2+r^2} - q\sqrt{x^2+y^2} &= 0,\tag1 \\
y\sqrt{q^2+r^2} - r\sqrt{x^2+y^2} &= 0.\tag2
\end{align}
We have to assume $c\neq 0,$ otherwise your equation is true everywhere.
So we can divide this into two cases.
In the first case, $r = 0$ but $q \neq 0.$ In that case,
Equation $(2)$ says that $y = 0,$ and therefore Equation $(1)$ says that
$x\lvert q\rvert = q \lvert x\rvert.$
This is true for all non-negative $x$ when $q > 0$ and for all non-positive $x$ when $q < 0.$
In the second case, $r\neq 0.$In that case,
Equation $(2)$ says that
$$ \sqrt{x^2+y^2} = \frac yr \sqrt{q^2+r^2}, $$
and plugging that into Equation $(1)$ we get
$$
x\sqrt{q^2+r^2} - y \frac qr \sqrt{q^2+r^2} = 0.
$$
This is an equation of the form $ax^2+ay^2+2(a'x+b'y)+c = 0$
where $a = 0,$ $a'= \frac12\sqrt{q^2+r^2},$ $b' = -\frac q{2r} \sqrt{q^2+r^2},$
and $c = 0.$
And it would be an equation of a line through the origin, except that
Equation $(2)$ also implies that $y$ has the same sign as $r,$ and therefore you only get half of the line.
Starting over from the beginning, a general equation for an arbitrary line in $\mathbb R^2$ (expressed in $(x,y)$ coordinates) is
$$Ax + By + C = 0, \tag3$$
where the coefficients $A,$ $B,$ and $C$ are whatever they need to be in order to produce an equation of the desired line.
An equation of this form exists for any line at any angle (even parallel to the $y$ axis) through any point in the plane.
For example:
To represent the line with equation $y = 3x + 5,$
simply set $A=3,$ $B=-1,$ and $C=5,$ and then you have the equivalent equation $3x + (-1)y + 5 = 0$, which is in the form $Ax + By + C = 0.$
To represent the line $y = -7,$ we can set $A=0,$ $B=1,$ and $C=7$ to get the equivalent equation $0x + 1y + 7 = 0.$
To represent the line $x = 15,$ simply set $A=1,$ $B=0,$ and $C=-15$ so that the equation $Ax + By + C = 0$ becomes $1x + 0y + (- 15) = 0.$
In short, the equation $Ax + By + C = 0$ is a very handy one (not just for this problem!) and it is good to make it as familiar as equations such as
$y = mx + k$ and $x^2 + y^2 = r^2.$
Surely it is not hard to see how to put the equation
$ax^2+ay^2+2(a'x+b'y)+c = 0$
into the form of Equation $(3).$
Best Answer
You must consider what is meant by $\sqrt\;$. There are several possible square root functions here: one has domain $\mathbb R^+$ and codomain $\mathbb R$, one has domain $\mathbb R$ and codomain $\mathbb C$, and one has domain $\mathbb C$ and codomain $\mathbb C$.
If you use the first function, then it is undefined when $z=\pm2$, so there is no solution.
If you use the second function, it is defined everywhere and $z=\pm2\in\mathbb R$ are solutions.
If you use the third function, it is defined everywhere and $z=\pm2\in\mathbb C$ are solutions.