<s>[INST] Determine all complex number z which satistfy the equation $|z + 1| = |z - i|$ I am trying to solve the aforementioned problem but have previously only worked with complex number inequalities where the resulting set is (the inside/outside of) a circle.
I am asked to determine all complex numbers z which satisfy the equation
$$|z + 1| = |z - i|$$
I have tried writing $z$ as $x + iy$ and grouping the real and imaginary parts together:
$|x + iy + 1| = |x + iy - i|$
but this just leads to
$|(x + 1) + i(y - 0)| = |(x + 0) + i(y - 1)|$
$\sqrt{(x + 1)^{2} + (y - 0)^{2}} = \sqrt{(x + 0)^{2} + (y - 1)^{2}}$
$(x + 1)^{2} + (y - 0)^{2} = (x + 0)^{2} + (y - 1)^{2}$
$x^2 + 2x + 1 + y^2 = x^2 + y^2 - 2y + 1$
$x + y = 0$
 [/INST] Let $z=x+iy$ with $x, y\,\in\,\bar{\mathbb{R}}$. We have:
$$|(x+1)+iy|=|x+i(y-1)|$$
So, using the definition of modulus:
$$\sqrt{(x+1)^2+y^2}=\sqrt{x^2+(y-1)^2}\implies x^2+2x+1+y^2=x^2+y^2-2y+1\implies 2x+1=-2y+1\implies x+y=0$$
Simply, we have a line that is the bisector of II and IV quadrant.
Also, there is a very nice geometrical interpretation, shown in this graph:

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