Solutions to $|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|$ Find all the complex numbers $z$ satisfying
\begin{equation}
|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|
\end{equation}
I tried using $z=a+bi$, then using the formula for absolute value:
\begin{equation}
|(a+1)+i(b-\sqrt{3})|=|(a-1)+i(b+\sqrt{3})|\\
\sqrt{(a+1)^2+(b-\sqrt{3})^2}=\sqrt{(a-1)^2+(b+\sqrt{3})^2}
\end{equation}
Is this the right way to go? Should I solve for b or a, and in that case what does that tell me about possible solutions? Any help would be appreciated.
Update:
Solved the equation for $y$, got
\begin{equation}
y=\frac{x}{\sqrt{3}}
\end{equation}
Does this mean that the equation is satisfied for all complex numbers on the line $y=\frac{x}{\sqrt{3}}$?
 A: Yes, this is a correct way to approach the problem. By squaring each side of your last line and simplifying, one obtains:
$$ 4a - 4\sqrt{3}b = 0 \Rightarrow a = \sqrt{3} b.$$
Thus the set that you are looking for consists of all the complex numbers $z = a + bi$ such that $a = \sqrt{3}b$, which in the complex plane is a straight line.
This was to be expected, because the equation can be rewritten as
$$ |z - (-1 + i\sqrt{3})| = |z - (1 - i\sqrt{3})|,$$
from where the problem can be solved geometrically: you are looking for points of the complex plane at an equal distance from $(-1 + i\sqrt{3})$ and $(1-i\sqrt{3})$.
A: Write 
$|z-(-1+i\sqrt{3})|=|z-(1-i\sqrt{3})|$
This is the locus of all $z$ which thier diatance from two points $A=-1+i\sqrt{3}$ and $B=1-i\sqrt{3}$ is equal. It halfs the fragment $AB$ which  should be $y=\sqrt{3}x$.
A: Hint-
You are on right way. Square both sides to find a, b. 
A: Hint: Square the equation and rewrite the square of the absolute values as the product of the complex number and its complex conjugate.
For example let $w=1-i\sqrt{3}$ then we can rewrite:
$|z+w|=|z-w| \implies |z+w|^2=|z-w|^2 \implies (z+w)(z^*+w^*)=(z-w)(z^*-w^*).$
This can be rewritten as:
$$zz^*+wz^*+zw^*+ww^*=zz^*-wz^*-zw^*+ww^*$$
$$wz^*+zw^*=-wz^*-zw^* \implies 2wz^*=-2zw^* \implies wz^*=zw^*$$
Now plug in $z = x+iy$ and $w=1-i\sqrt{3}$. It should be easy to solve from here.
A: $$
|z + 1 - i\sqrt{3}| = |z - 1 + i\sqrt{3}| \\
\implies |z - (-1 + i\sqrt{3})| = |z - (1 - i\sqrt{3})|
$$
If I take $z_0 = 1 - i\sqrt{3}$, then the equation
$$
|z - (-z_0)| = |z - z_0|
$$
describes the perpendicular bisector of the straight line joining the points $P(z_0)$ and $Q(-z_0)$. What is clear is that the required line passes through the origin. The slope of $PQ$ is
$$
m_{PQ} = \frac{\sqrt{3} + \sqrt{3}}{-1-1} = -\sqrt{3}.
$$
Hence a line perpendicular to $PQ$ has a slope $1/\sqrt{3}$. The required line is thus
$$
y = \frac{x}{\sqrt{3}} \\
\implies \sqrt{3}y = x.
$$
