STEP II 1995 question 6 solution - The Student Room
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STEP II 1995 question 6 solution

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 (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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