Let z1-1-i and z2-1-i- Find z3 - C such that triangle z1z2z3 is equilateral-

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Perpendicular Distance of a Point from a Plane

let z1=1+i ...

Question

let $z_{1} = 1 + i$ and $z_{2} = - 1 - i$ . Find $z_{3} \in C$ such that triangle $z_{1} z_{2} z_{3}$ is equilateral.

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Solution

Given,
$z_{1}, z_{2}, z_{3}$ are three complex numbers joined to form an equilateral triangle the coordinates of $z_{1}$ are $(1, 1)$
and that of $, z_{2}$ are $(- 1, 1)$ .
and that of $z_{3}$ are $(x, y)$ (say)
There's no harm in considering complex numbers as coordinates in a real plane.After all, there isn't much distinction while representing graphically.Now using the distance formula,we can write
${(x - 1)}^{2} + {(y - 1)}^{2} = {(x + 1)}^{2} + {(y + 1)}^{2} = 8$
Solving we get
$x + y = 1 - (i) x^{2} + y^{2} = 8 - 2 = 6 - (i i)$
Solving for $x$ & $y$ we get
$x = \pm \sqrt{3} y = \pm \sqrt{3}$
$∴ z_{3} = (\sqrt{3} + i \sqrt{3})$ or $(\sqrt{3} - i \sqrt{3})$

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let z1=1+i and z2=−1−i. Find z3∈C such that triangle z1z2z3 is equilateral.

let $z_{1} = 1 + i$ and $z_{2} = - 1 - i$ . Find $z_{3} \in C$ such that triangle $z_{1} z_{2} z_{3}$ is equilateral.