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

$$
|z + 1 - i\sqrt{3}| = |z - 1 + i\sqrt{3}| \\
\implies |z - (-1 + i\sqrt{3})| = |z - (1 - i\sqrt{3})|
$$
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
$$
y = \frac{x}{\sqrt{3}} \\
\implies \sqrt{3}y = x.
$$
