Drawing $\lvert z-2\rvert + \lvert z+2\rvert=5$ in complex plane If we substitute $z=x+iy$ to $\lvert z-2\rvert + \lvert z+2\rvert=5$ and solve for $iy$ we will get
$$ iy=-\frac{1}{2}(2x-5) $$
$$ iy=-\frac{1}{2}(2x+5) $$
Then we can draw like in the $\mathbb{R}^2$ plane. This way we get two parallel lines.
Do you think this approach is correct for the complex plane? Just want to make sure.

$1)$ If you want go through a geometric approach: by definition of ellipse one can conclude that $z$ is on an ellipse with foci $-2$ and $2$ and bigger axis equal to $5$.
$2)$ If you want go through an analytic approach you can write $z=x+iy$ and then
$$\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=5$$
$$\sqrt{(x-2)^2+y^2}=5-\sqrt{(x+2)^2+y^2}$$
Square both sides:
$$(x-2)^2+y^2=25-10\sqrt{(x+2)^2+y^2}+(x+2)^2+y^2$$
$$10\sqrt{(x+2)^2+y^2}=25+8x \rightarrow 100[(x+2)^2+y^2]=(25+8x)^2$$
Can you finish?
