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Let $a, b, c$ be distinct complex numbers such that $\frac{a}{1-b}=\frac{b}{1-c}=\frac{c}{1-a}=k$. Find the value of $k$.

My approach:
$a+b+c=k(3-(a+b+c))$, after this step not able to proceed

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1 Answer 1

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$$a=k-kb,b=k-kc,c=k-ka$$ $$a-b=-k(b-c),b-c=-k(c-a),c-a=-k(a-b)$$ $$a-b=-k^3(a-b)\implies k^3=-1\implies k=-1,e^{i\pi/3},e^{-i\pi/3}$$

If $k=-1$ then $a+b+c=3+a+b+c$, contradiction.

The other two values of $k$ have solutions, for example $(a,b,c)=(\frac{2k}{k+1},\frac{k-1}{k+1},\frac1{k+1})$

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