Solutions to $|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|$

Question

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

Answer 1

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

Answer 2

Hint-

You are on right way. Square both sides to find a, b.

Answer 3

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.

Answer 4

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 $y=\sqrt{3}x$.

Answer 5

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

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

$$ y = \frac{x}{\sqrt{3}} \\ \implies \sqrt{3}y = x. $$