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Question

If a = cos(2pi/7)+isin(2pi/7), L = a+a2+aand M = a3+a5+a6, then L,M are roots of the equation _________


Solution

 

This is a bit tricky question. The idea behind the solution which I give you here can be used in many places to get the quadratic equation, whose roots satisfy some given conditions.

First of all: a^0 + a^1 +......+ a^6 = (a^7 - 1)/(a - 1) [using the sum of GP formula]

But a^7 = cos(2pi) + i*sin(2pi) = 1..... Hence the above sum = 0.

For the sake of typing, let me use "p" in place of alpha and "q" in place of beta.

So sum of roots = p+q = a + a^2 +.....a^6 = -a^0 = -1.

Next p^2 = (a + a^2 + a^4)^2 = a^2 + a^4 + a^8 + 2(a^3 + a^5 + a^6) ---------------- (1)

Also note that a^8 = a^7*a = a [since a^7 = 1]

Hence you have p^2 = p + 2(q) --------------[from (1)]

But we also have shown that p+q = -1, so q = -1-p

Hence p^2 = p -2*(1+p)

ie p^2 + p + 2 = 0......... (similarly you can also check up that q^2 + q + 2 = 0, by substitution of p = -1-q in the equation)

Hence both p & q satisfy the equation, x^2 + x + 2 = 0 (where p & q are the roots of that equation).....

And that solves the problem.




 


Mathematics

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