$z_1, z_2, z_3\in\mathbb C$ satisfy $|z_1-3|=|z_2-3|=|z_3-3|$ and $\arg(\frac{z_3-z_1}{z_2-z_1})=\pi/6$, compute $z_2^2+z_3^2-3z_2-3z_3-z_2z_3+10$

Question

let $z_{1}, z_{2}, z_{3}$ be three complex number such that $|z_{1}-3|=|z_{2}-3|=|z_{3}-3|$ and $\arg(\frac{z_{3}-z_{1}}{z_{2}-z_{1}})=\pi/6$

then what is the value of $z_{2}^2+z_{3}^2-3z_{2}-3z_{3}-z_{2}*z_{3}+10$

my approach:

I identified that the three complex numbers lie on a circle of variable radius and given center and the argument relation gave use difference between angles subtend by any two of them from origin but I couldn't use this to create the relation of the value they have asked

Answer 1

Let $w_1=z_1-3, w_2=z_2-3, w_3=z_3-3$, then $$|w_1|=|w_2|=|w_3|=:r>0,\qquad \arg\left(\frac{w_3-w_1}{w_2-w_1}\right)=\frac\pi6,$$ and \begin{align*} z_{2}^2+z_{3}^2-3z_{2}-3z_{3}-z_{2}z_{3}+10=w_2^2+w_3^2-w_2w_3+1. \end{align*}

We use the geometric meaning of complex numbers. Draw a circle $C$ in the $xOy$ plane centering at the origin with raduis $r$. Let $P, Q, R$ denote $w_1, w_2, w_3$ respectively, then $w_3-w_1$ corresponds to the vector $\vec{PR}$ and $w_2-w_1$ corresponds to the vector $\vec{PQ}$. The condition $\arg\left(\frac{w_3-w_1}{w_2-w_1}\right)=\frac\pi6$ implies that $\angle QPR=\frac\pi6$ and $P, Q, R$ lies in a counterclockwise manner. Hence $\angle QOR=\frac\pi3$ and thus $w_3=\eta w_2$ with $\eta=e^{\frac\pi3i}=\frac12+\frac{\sqrt3}2i$. Therefore, $$w_2^2+w_3^2-w_2w_3=(\eta^2-\eta+1)w_2^2=0.$$ Hence $$z_{2}^2+z_{3}^2-3z_{2}-3z_{3}-z_{2}z_{3}+10=w_2^2+w_3^2-w_2w_3+1=1.$$

Answer 2

This implies that triangle formed by $3, z_2, z_3$ as vertices is equilateral

It is well known that $a,b,c$ represent vertices of an equilateral triangle iff

$a^2+b^2+c^2-ab-bc-ca =0$ i.e. $z_2^2+z_3^2+9 - 3z_2-3z_3-z_2z_3 = 0$ so that given expression equals $1$