$$|z+1+3i|=|z-5-7i|$$ $z$ represents a complex number right? Then if $$|z+1+3i|=0$$ $${\implies}|z|=|-1-3i|$$ In which sense does this $$|z+1+3i|=|z-5-7i|$$ imply, $$\implies|-4-4i|$$ But $z$ has vanished, which begs the question what is the point of including $z$ at all?
I know I'm getting something wrong, but what? Regards Tom
On
The notation should be interpreted as magnitude, not absolute value signs.
If we let $z$, the complex number, be represented as $z = a + bi$, we can then say that:
$$\left|a+1 + i(b+ 3)\right| = \sqrt{(a+1)^2 + (b+3)^2}$$
and that:
$$\left|a-5 + i(b-7)\right| = \sqrt{(a-5)^2 + (b-7)^2}$$
Setting them equal to one another, we get that:
$$\sqrt{(a+1)^2 + (b+3)^2} = \sqrt{(a-5)^2 + (b-7)^2}$$
$$(a+1)^2 + (b+3)^2 = (a-5)^2 + (b-7)^2 $$
$$a^2 + 2a + 1 + b^2 + 6b + 9 = a^2 - 10a + 25 + b^2 -14b + 49$$
$$12a + 20b = 64,\ 3a + 5b = 16$$
One possible solution is $a = -3$, $b = 5$. You can derive a set of solutions that satisfy this condition using the Euclidean Algorithm.
Therefore, $z$ can be: $z = -3 + 5i$
It has nothing to see with $\lvert z+1+3\mathrm i\rvert=0$. Geometrically, the equation means the image of $z$ is at the same distance from the image $A$ of $-1-3\mathrm i$ as from the image $B$ of $5+7\mathrm i$.
The locus of such points is the midperpendicular of $[AB]$; the midpoint of $[AB]$ is the point with affix $\frac12(4+4\mathrm i)$.
Now the line $(AB)$ is directed by the vector $(3,5)$, hence perpendicular is directed by the vector $(-5,3)$, with affix $-5+3\mathrm i$. A point of the midperpendicular is $(2,2)+t(-5,3)(2-5t,2+3t),\enspace t\in\mathbf R$. Thus the solutions are: $$z=2-5t+(2+3t)\mathrm i.$$