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

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 $(1-i\sqrt{3})$.
