Question
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}
Answer
Let \dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=k, then
a = bk
(1)
c=ak=bk^2
(2)
b=ck=bk^3
(3)
Then
k^3=1
(4)
Apparantly k=1 is one of the roots for the equation. To cover all cases, solving the cubic equation gives another two complex roots.
(k-1)(k^2+k+1)=0
\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
\dfrac{a+b-c}{a-b+c}=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}