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STEP III 1997 question 3 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1997 question 3 solution


The roots of the polynomial z^n - 1 are 1, \omega, \omega^2, \ldots and hence the polynomial can be written.

(z - 1)(z - \omega)(z-\omega^2)\cdots(z -\omega^{n-1}) = z^n - 1.

Dividing by z - 1, and converting the formula for the sum of a geometric sequence into and actual sum of a geometric sequence, we get

(z - \omega)(z - \omega^2)\cdots(z-\omega^{n-1}) = 1 + z + z^2 + z^3 + \cdots + z^{n-1}

as required.

Next part: Embedding the problem in the Argand plane, we can let A_1, A_2, \ldots, A_n be represented by the complex numbers r, r\omega, r\omega^2, \ldots, r\omega^{n-1}, respectively. the point O then corresponds to the complex number 0.

But then, let z = \omega in the above proved formula. Then, the LHS becomes zero, and the sum in the RHS is precisely the sum of the position vectors of A_1, A_2, \ldots, A_n, which must therefore also be zero, as required.

Let the point O have coordinates (0, 0), the point A_1 have coordinates (r, 0).

Then the coordinates of A_k for  1 \leq k \leq n are given by (r\cos(k\pi/n), r\sin(k\pi/n)).

Now we have

\displaystyle \sum_{k=1}^n |A_1A_k|^2

\displaystyle = \sum_{k=1}^n \left((r\cos(k\pi/n) - r)^2 + (r\sin(k\pi/n))^2\right)

\displaystyle = \sum_{k=1}^n \left(2r^2 - 2r^2\cos(k\pi/n)\right)

But

\displaystyle \sum_{k=1}^n r\cos(k\pi/n) = 0

(This is basically the horizontal bit of the vector identity proved in the last part)

So

\displaystyle \sum_{k=1}^n |A_1A_k|^2

\displaystyle = \sum_{k=1}^n 2r^2

\displaystyle = 2r^2n

as required.

Solution by ukgea

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