STEP II 1995 question 6 solution

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STEP II 1995 question 6 solution

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TSR Wiki]] > Study Help > Exams and Qualifications > A Levels > STEP > STEP II 1995 question 6 solution

$(z - u)(z - v) = z^{2} - (u + v) z + uv \equiv z^{2} + az + b \implies a = -u - v, \ b = uv$ .

$\alpha = \cos{ \frac{ 2 \pi }{7} } + i \sin{ \frac{ 2 \pi }{7} } = e^{ \frac{ 2 \pi }{7} i } \implies \alpha ^{7} = e^{ 2 \pi i } = 1$ .

The other 7th roots of unity are the roots of the equation:

$\alpha ^{2} , \ \alpha ^{3} , \ \alpha ^{4} , \ \alpha ^{5} , \ \alpha ^{6} , \ 1$ .

$u = \alpha + \alpha ^{2} + \alpha ^{4}$

$v = \alpha ^{3} + \alpha ^{5} + \alpha ^{6}$ .

Hence:

$A = - u - v = - (u + v) = - \left( \alpha + \alpha ^{2} + \alpha ^{3} + \alpha ^{4} + \alpha ^{5} + \alpha ^{6} \right) = - \left( -1 \right) = 1$

$B = uv = \left( \alpha + \alpha ^{2} + \alpha ^{4} \right) \left( \alpha ^{3} + \alpha ^{5} + \alpha ^{6} \right) = 3 + \alpha ^{4} + \alpha ^{6} + \alpha ^{5} + \alpha + \alpha ^{2} + \alpha ^{3} = 3 - 1 = 2$ .

Hence:

$z^{2} + z + 2 = 0$ .

Hence:

$z = u , \ or \ v = \frac{ -1 \pm \sqrt{ -7 }}{2} = \frac{ -1 }{2} \pm i \frac{ \sqrt{7} }{2}$ .

Consider:

$Re \left( \alpha + \alpha ^{2} + \alpha ^{4} \right) = \cos{ \frac{ 2 \pi }{7} } + \cos{ \frac{ 4 \pi }{7} } + \cos{ \frac{ 8 \pi }{7} }$ .

Hence:

$\cos{ \frac{ 2 \pi }{7} } + \cos{ \frac{ 4 \pi }{7} } + \cos{ \frac{ 8 \pi }{7} } = Re(u) = \frac{ -1 }{2}$ .

$Im \left( \alpha + \alpha ^{2} + \alpha ^{4} \right) = \sin{ \frac{ 2 \pi }{7} } + \sin{ \frac{ 4 \pi }{7} } + \sin{ \frac{ 8 \pi }{7} }$ .

The function of "x", $f(x) = \sin{x}$ is increasing from x = 0 to $x = \frac{ \pi }{2}$ . It is positive from x = 0 to $x = \pi$ , and negative from $x = \pi$ to $x = 2 \pi$ (the behaviour after is not relevant to this question).

$\sin{ \frac{ 8 \pi }{7} } = \sin{ \left( \pi - \frac{ 8 \pi }{7} \right) } = \sin{ \frac{ - \pi }{7} } = - \sin{ \frac{ \pi }{7} }$ .

$\sin{ \frac{ \pi }{7} } < \sin{ \frac{ 2 \pi }{7} }$

Hence one wishes to have the positive value for the aforementioned "imaginary part":

$Im \left( \alpha + \alpha ^{2} + \alpha ^{4} \right) = \sin{ \frac{ 2 \pi }{7} } + \sin{ \frac{ 4 \pi }{7} } + \sin{ \frac{ 8 \pi }{7} } = \frac{ \sqrt{7}}{2}$ .

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