$|z+2|=|z|-2$; Represent on an Argand Diagram

Question

Represent on an Argand Diagram the set given by the equation $|z+2|=|z|-2.$

My attempt:
Apparently the answer is $x\leq 0$ $(z = x + yi)$ and $y = 0$, based on the idea that $-x = \sqrt{(x^2 + y^2)}$, but I am struggling to derive this.

I originally assumed the answer was $y = 0, x\leq-2$, going from the idea that the distance of $z$ from $(-2,0)$ is the same as the distance from the origin minus $2.$ However, this is not the solution.

Any help would be much appreciated

Answer 1

The solution given to you was wrong. Notice, for instance, if $x=y=0$, then $z=0$ but $z=0$ leads to $|0+2| =|0|-2 \implies 2 = -2$. Obviously, this is nonsense.

Indeed, your reasoning is correct. Through simple algebraic manipulation you can conclude $y=0$, and from there you can reduce the equation to $|x+2| = |x|-2$. Simply graphing it alone shows that $f(x) = |x+2| - |x| + 2$ satisfies $f(x) = 0$ (i.e. the desired $x$ values) only when $x \le -2$:

Answer 2

The equation $\,|z+2|=|z|-2\,$ is equivalent to $|z|-|z+2|=2.$

With a value $0<c<2$ on the right side would the equation $$|z|-|z+2|=c$$ define one branch of hyperbola with foci at $0$ and $-2,$ the branch near the focus at $-2.$

As $\,c \to 2,\,$ the hyperbola is thinner and thinner, as shows the picture.
For $\,c=2\,$ it collapses into a half-line (red) defined as

$$\Re (z) \leq -2 \quad \text{and} \quad \Im (z) =0.$$

Answer 3

Note that $|z+2|=|z|-2\ge 0$ implies $|z| \ge 2$ and

\begin{align} & (z+2)(\bar z +2) = z\bar z - 4|z| +4 \\ \implies & z+\bar z =- 2 |z| \le 0\\ \implies & (z+\bar z)(\bar z +z) = 4 z\bar z \\ \implies & (z-\bar z)^2 =0 \\ \end{align}

Thus, with $z=x+i y$, we have $y=\frac1{2i}(z-\bar z) =0$ and, since $z+\bar z<0,\> |z|\le 2$, we also have $x= \frac12(z+\bar z) = -|z| \le 2$

Answer 4

Start with $$ |z+2|=|z|-2\tag1 $$ square both sides using $|z+2|^2=(z+2)(\bar z+2)=z\bar z+2(z+\bar z)+4$, we get $$ |z|^2+4\mathrm{Re}(z)+4=|z|^2-4|z|+4\tag2 $$ which gives $$ \mathrm{Re}(z)=-|z|\tag3 $$ squaring $(3)$ gives $$ \mathrm{Re}(z)^2=|z|^2=\mathrm{Re}(z)^2+\mathrm{Im}(z)^2\tag4 $$ $(4)$ says that $\mathrm{Im}(z)=0$ and $(3)$ says that $\mathrm{Re}(z)\le0$. Therefore, $z$ is a negative real, which means that $|z|=-z$. Thus, $(1)$ becomes $$ |z+2|=-(z+2)\tag5 $$ which is true precisely when $z+2\le0$. Therefore, we get that $$ \mathrm{Im}(z)=0\text{ and }\mathrm{Re}(z)\le-2\tag6 $$ Thus, your original idea was correct.