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 $y=\frac{x}{\sqrt{3}}$?

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 $y=\sqrt{3}x$.
