If a,b,c are nonzero complex numbers such that \dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}, find all possible values for \dfrac{a+b-c}{a-b+c}

Collected in the board: Imaginary number

Steven Zheng posted 1 week ago


Let \dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=k, then

a = bk



Apparantly k=1 is one of the roots for the equation. To cover all cases, solving the cubic equation gives another two complex roots.


\begin{cases} k_1=1 \\ k_2=\dfrac{-1+\sqrt{3}i }{2} \\ k_3=\dfrac{-1-\sqrt{3}i }{2} \end{cases}

in which k_2 and k_3 are also denoted as ω and \overline{ω}. Apparently, ω^2 = \overline{ω}, and vice versa.

Substitute the value of k to (1), (2), (3) to obtain the value of a,b,c, and then determine the value of \dfrac{a+b-c}{a-b+c} in three cases depending the values of k.

Case 1, if k = 1


Case 2, if k=ω

\dfrac{a+b-c}{a-b+c}=\dfrac{bω+b-bω^2}{bω-b+bω^2} =\dfrac{ω+1-\overline{ω}}{ω-1+\overline{ω}}=-\dfrac{1+\sqrt{3}i }{2}

Case 3, if k=\overline{ω}

\dfrac{a+b-c}{a-b+c}=\dfrac{b\overline{ω}+b-b\overline{ω}^2}{b\overline{ω}-b+b\overline{ω}} =\dfrac{\overline{ω}+1-ω}{\overline{ω}-1+ω}=-\dfrac{1-\sqrt{3}i }{2}

Steven Zheng posted 1 week ago

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