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Given three complex numbers $a,b,c$ such that $\lvert a \rvert=\lvert b \rvert=\lvert c \rvert=1$ and $$\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=-1$$

Find which of the following are possible values of $\lvert a+b+c \rvert$

A)0

B)2

C)1.5

D)3

My try:

I assumed $a=e^{ix}$,$b=e^{iy}$, $c=e^{iz}$

Then we have

$$\cos (2x-y-z)+\cos (2y-x-z)+\cos (2z-x-y)=-1$$

$$\sin(2x-y-z)+\sin (2y-x-z)+\sin (2z-x-y)=0$$

Squaring and adding we get

$$3+2(\cos(3x-3y)+\cos(3y-3z)+\cos (3z-3x)=1$$

So

$$\cos(3x-3y)+\cos(3y-3z)+\cos (3z-3x)=-1$$

any clue here?

Ekaveera Gouribhatla
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2 Answers2

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We will use some facts such as $|a|=1$ implies $a\bar{a}=1$. So $\frac{1}{a}=\bar{a}$. Also let $w=a+b+c$, then $|w|^2=(a+b+c)(\bar{a}+\bar{b}+\bar{c})$.

\begin{align*} \frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}&=-1\\ \frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}&=-1\\ a^3+b^3+c^3+abc & =0\\ a^3+b^3+c^3-3abc & =-4abc\\ (a+b+c)(a^2+b^2+c^2-ab-bc-ca)&=-4abc\\ (a+b+c)((a+b+c)^2-3ab-3bc-3ca)&=-4abc. \end{align*} we get \begin{align*} w^3-3w(ab+bc+ca)&=-4abc\\ w^3-3wabc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)&=-4abc\\ w^3&=abc\left[3w\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-4\right]\\ w^3&=abc\left[3w\left(\bar{a}+\bar{b}+\bar{c}\right)-4\right]\\ w^3&=abc\left[3|w|^2-4\right]\\ |w^3|&=\left|3|w|^2-4\right|\\ \tag{1} \label{eq1} |w|^3&=\left|3|w|^2-4\right|\\ \end{align*}

  1. Case 1: If $3|w|^2-4 \geq 0$, then \eqref{eq1} becomes $|w|^3-3|w|^2+4=0$ and this implies $|w|=2$.
  2. Case 2: If $3|w|^2-4 < 0$, then \eqref{eq1} becomes $|w|^3+3|w|^2-4=0$ and this implies $|w|=1$.

Thus $|w| \in \{1,2\}$.

Anurag A
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The four complex numbers$\frac{a^2}{bc},\frac{b^2}{ac},\frac{c^2}{ab},1$ all have length $1$ and sum to zero. Added as vectors,'head to tail', in the Argand diagram they form a closed shape and must consist of two pairs of equal but opposite vectors.

Without loss of generality we therefore have $\frac{a^2}{bc}=-\frac{b^2}{ac},\frac{c^2}{ab}=-1$ and then $a^3=-b^3,ab=-c^2$. Write $c$ as $wa$, then $b=-w^2a$, where $w$ must be a 6th root of unity.

Then $|a+b+c|=|1+w-w^2|$ is $2$ if $w$ is a primitive 3rd or 6th root and is $1$ if $w$ is $1$ or $-1.$ The answer is B.