How to solve this absolute value equation: $|z-3|-|z+6|=0,z\in\mathbb C$

Question

So I was bored, and decided to solve some absolute value equations. After a while, I came up with this$$|z-3|-|z+6|=0,z\in\mathbb C$$which I thought that I might be able to solve. Here is my attempt at doing so:

---

For any absolute value of a complex number, we have$$|a|\implies\sqrt{a^2}\\|a+bi|\implies\sqrt{a^2+b^2}\quad a,b\in\mathbb R\\\implies|a+bi+c|=\sqrt{a^2+2ac+c^2+b^2}\quad a,b,c\in\mathbb R$$So we now have the equation$$\sqrt{a^2-6a+9+b^2}-\sqrt{a^2+12a+36+b^2}=0\\\implies a^2-a^2+b^2-b^2-6a-12a+9-36=0\\\implies-27-18a=0\implies18a+27=0\implies a=-\dfrac{27}{18}$$Now to solve for $b$. So we have$$\sqrt{\dfrac{729}{324}-\dfrac{162}{18}+9+b^2}=\sqrt{\dfrac{729}{324}+54+b^2}\\\implies\sqrt{\dfrac94+b^2}=\sqrt{\dfrac{225}4+b^2}\\\implies\dfrac94+b^2=\dfrac{225}4+b^2$$meaning that for no $bi$ will make this statement true, meaning that our solution is$$z=-\dfrac{27}{18}+0i$$which we can check in Desmos to confirm our answer.

---

My question

---

Is it true that the only solution to$$|z-3|-|z+6|=0,z\in\mathbb C$$is $z=-\dfrac{27}{18}+0i$, or what might I have done wrong?

Answer 1

Geometrically, $|z-3|$ is the distance of $z$ from the point $3$ in the complex plane. Similarly, $|z+6|$ is the distance of $z$ from the point $-6$. So $|z-3| - |z+6| = 0$, which is equivalent to $|z-3| = |z+6|$, means that the distance of $z$ from $3$ is equal to the distance of $z$ from $-6$. Clearly, the locus of points in a plane that are equidistant from two fixed points is the perpendicular bisector of the line segment joining the two points.

So how would we find this line using Cartesian geometry? We know that $3 = (3,0)$ and $-6 = (-6,0)$. So the midpoint is $(-3/2, 0)$ and the perpendicular bisector would simply be $x = -3/2$ since the $y$-coordinates of the two points are the same. In the complex plane, this means $$z = -\frac{3}{2} + yi,$$ for all real $y$.

Now how do we get this result algebraically from the original equation? We would write $z = x+yi$, hence

$$|z-3| = |(x-3)+yi| = \sqrt{(x-3)^2 + y^2},$$ and similarly, $$|z+6| = \sqrt{(x+6)^2 + y^2}.$$ Hence $|z-3| = |z+6|$ implies $$\sqrt{(x-3)^2 + y^2} = \sqrt{(x+6)^2 + y^2}.$$ Squaring both sides, $$(x-3)^2 = (x+6)^2,$$ hence $$x^2-6x+9 = x^2+12x+36$$ or $$x = -\frac{3}{2}.$$ The value of $y$ doesn't matter; any real $y$ will satisfy the conditions, which you can check by substituting $z = -\frac{3}{2} + yi$ into the original equation.

Answer 2

Let$$z=x+iy$$Now according to your question $$|z-3|-|z+6|=0$$ we have $$|(x-3)+iy|-|(x+6)+iy|=0$$Now, according to the concept of complex numbers, $$|x+iy|=(x^2+y^2)^{1/2}$$ So, the final equation will be $$(x-3)^2+y^2=(x+6)^2+y^2$$. Now solve $$(x-3)^2=(x+6)^2$$ So, $x$ will be equal to $-27/18$. So, $z=(-27/18)+0i$. Although, I am not sure whether my answer is correct or not. If someone would please check it that would be nice.

Answer 3

I would recommend you to notice that:

\begin{align*} |z - 3| - |z + 6| = 0 & \Longleftrightarrow |z - 3| = |z + 6|\\\\ & \Longleftrightarrow |z - 3|^{2} = |z + 6|^{2}\\\\ & \Longleftrightarrow (z - 3)(\overline{z} - 3) = (z + 6)(\overline{z} + 6)\\\\ & \Longleftrightarrow z\overline{z} - 3(z + \overline{z}) + 9 = z\overline{z} + 6(z + \overline{z}) + 36\\\\ & \Longleftrightarrow 9(z + \overline{z}) = -27\\\\ & \Longleftrightarrow 18x = -27\\\\ & \Longleftrightarrow x = -\frac{3}{2} \end{align*} where $z = x + yi$.