Matematické Fórum

stuart clark · 26. 04. 2011 06:57

(4) If $a,b,c$ are distinct Complex no. such that $\displaystyle \frac{a}{1-b}=\frac{b}{1-c}=\frac{c}{1-a}=k$ .then find value of $k$

musixx · 28. 04. 2011 11:17 — Editoval musixx (28. 04. 2011 12:40)

Clearly, a, b, and c are distinct from 1, and if one of them is zero, then also all others together with k are zero (EDIT: In the end, that is a contradiction with the condition that a,b,c are distinct complex numbers).

So let's assume all of them to be distinct from 0 and 1, and let's assume that such k exists.

If we compute c from $\frac a{1-b}=\frac b{1-c}$ and from $\frac a{1-b}=\frac c{1-a}$ , and if we compare those two results, we have an equation $a^3-a^2+a(1-b)-b(1-b)^2=0$ . We are looking for $k=\frac a{1-b}$ , so let's divide that equation with nonzero $(1-b)^3$ , and let's introduce a common trick "+1-1" to a fraction $\frac b{1-b}$ . That transforms our equation to $k^3+1-\frac{k^2-k+1}{1-b}=0$ . Since $x^3+1=(x+1)(x^2-x+1)$ , we have $(k^2-k+1)$\frac{-b}{1-b}+k$=0$ .

But $\frac{-b}{1-b}+k=\frac{a-b}{1-b}$ , and we assume $a\neq b$ , so therefore $k^2-k+1=0$ which gives $k=\frac12\pm{\rm i}\frac{\sqrt3}2$ .

It is easy to verify that both solutions are possible.

stuart clark · 29. 04. 2011 08:32

Thanks musixx.

Matematické Fórum

#1 26. 04. 2011 06:57

Complex no.

#2 28. 04. 2011 11:17 — Editoval musixx (28. 04. 2011 12:40)

Re: Complex no.

#3 29. 04. 2011 08:32

Re: Complex no.

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