Complex equation solution. How can i resolve it? I have this complex equation $|z+2i|=| z-2 |$. How can i resolve it? Please help me
 A: The geometric way
The points $z$ that satisfy the equation are at the same distance of the points $2$ and $-2\,i$, that is, they are on the perpendicular bisector of the segment joining $2$ and $-2\,i$. This is a line, whose equation you should be able to find.
The algebraic way
When dealing vith equations with $|w|$, it is usually convenient to consider $|w|^2=w\,\bar w$. In your equation, if $z=x+y\,i$, this leads to $x^2+(y+2)^2=(x-2)^2+y^2$, whose solution I'll leave to you.
A: We have $|z+2i|^2=| z-2 |^2$, which implies that
$(z+2i)\overline{(z+2i)}=(z-2)\overline{(z-2)}$, that is
$(z+2i)(\overline{z}-2i)=(z-2)(\overline{z}-2)$. This implies that
$$z\overline{z}-2iz+2i\overline{z}+4=z\overline{z}-2z-2\overline{z}+4,$$
that is
$-iz+i\overline{z}=-z-\overline{z}$, or 
$$i(z-\overline{z})=z+\overline{z}.$$
Now if we write $z=a+bi$, we get
$i(2bi)=2a$, or $b=-a$.
A: Let $z=a+bi$ be a complex number that satisfies
$$|z+2i|=|z-2|.$$
Remember that for any complex number $c+di$, the definition of $|c+di|$ is
$$|c+di|=\sqrt{c^2+d^2}$$
So, we have that
$$|z+2i|=|(a+bi)+2i|=|a+(b+2)i|=\sqrt{a^2+(b+2)^2}$$
and
$$|z-2|=|(a+bi)-2|=|(a-2)+bi|=\sqrt{(a-2)^2+b^2}.$$
Are you able to solve for the values of $a$ and $b$ such that the two expressions above are equal?
