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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

$|(x + 1) + i(y - 0)| = |(x + 0) + i(y - 1)|$

$\sqrt{(x + 1)^{2} + (y - 0)^{2}} = \sqrt{(x + 0)^{2} + (y - 1)^{2}}$

$(x + 1)^{2} + (y - 0)^{2} = (x + 0)^{2} + (y - 1)^{2}$

$x^2 + 2x + 1 + y^2 = x^2 + y^2 - 2y + 1$

$x + y = 0$

Hector Auvinen
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    Your calculations are right and you basically answered the problem yourself. The condition $x+y=0$ just means that all complex numbers of the form $z= x-ix = x(1-i)$ satisfy the condition $|z+1|=|z-i|$ – Lukas Feb 09 '22 at 14:09

2 Answers2

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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:

enter image description here

Matteo
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  • @Hector Auvinen: if this answer has been useful, let me know smashing that +1 button or accepted answer. Thanks. – Matteo Feb 09 '22 at 18:44
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The points equidistant from two points in the plane - in this case, $-1,\,i$ - form a line perpendicular to the line segment joiing those points. Treating $\Bbb C$ as $\Bbb R^2$ in the usual way, the segmene is from $(-1,\,0)$ to $(0,\,1)$ with gradient $1$, so the desired locus is a line of gradient $-1$. Clearly $-1,\,0,\,i$ are the vertices of an isosceles right-angled triangle, so $0$ is on this new line. That gives it the equation $x+y=0$, no fiddling with square roots needed.

J.G.
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