Let $A(a),B(b),C(c)$ be vertices of a triangle in the complex plane. Does the following relation hold? \begin{align*} |a|^2-|b|^2+(\epsilon+1)(\overline{b}c-a\overline{c})+(\epsilon-2)(\overline{a}c-b\overline{c})=0 \end{align*} where $\epsilon=\frac{1}{2}+\frac{i\sqrt{3}}{2}$
Is this relation for complex numbers true?
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complex-numbers
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0Note that since you say nothing about properties of this triangle, $a,b,c$ are really just arbitrary complex numbers. (Arguably the condition implies that $a,b,c$ are not collinear, but your expression is a polynomial in the real and imaginary parts of $a,b,c$, so this doesn't matter.) – 2017-01-13
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0Now I see, thank you! – 2017-01-13
1 Answers
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No. For $a=i$, $b=-i$ and $c=1-i$ this expression is false. This relation between $a$, $b$ and $c$ is equivalent with $${\bf Re}\Big(|a|^2-|b|^2+2\delta(\bar{b}c-a\bar{c})\Big)=0$$ where $\delta=\epsilon+1$.