OCR Further Pure Core 2 2024 June — Question 9

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2024
SessionJune
TopicComplex numbers 2

9 In this question, the argument of a complex number is defined as being in the range \([ 0,2 \pi )\).
You are given that \(\omega _ { k }\), where \(k = 0,1,2 , \ldots , n - 1\), are the \(n n ^ { \text {th } }\) roots of unity for some integer \(n , n \geqslant 3\), and that these are given in order of increasing argument (so that \(\omega _ { 0 } = 1\) ).
  1. With the help of a diagram explain why \(\omega _ { k } = \left( \omega _ { 1 } \right) ^ { k }\) for \(k = 2 , \ldots , n - 1\).
  2. Using the identity given in part (a), show that \(\sum _ { \mathrm { k } = 0 } ^ { \mathrm { n } - 1 } \omega _ { \mathrm { k } } = 0\).
  3. Show that if \(z\) is a complex number then \(z + z ^ { * } = 2 \operatorname { Re } ( z )\).
  4. Using the results from parts (b) and (c) show that \(\sum _ { \mathrm { k } = 0 } ^ { \mathrm { n } - 1 } \operatorname { Re } \left( \omega _ { \mathrm { k } } \right) = 0\).
  5. With the help of a diagram explain why \(\operatorname { Re } \left( \omega _ { \mathrm { k } } \right) = \operatorname { Re } \left( \omega _ { \mathrm { n } - \mathrm { k } } \right)\) for \(k = 1,2 , \ldots , n - 1\). You should now consider the case where \(n = 5\).
    1. Use parts (d) and (e) to deduce that \(\cos \frac { 4 \pi } { 5 } = \mathrm { a } + \mathrm { b } \cos \frac { 2 \pi } { 5 }\), for some rational constants \(a\) and \(b\).
    2. Hence determine the exact value of \(\cos \frac { 2 \pi } { 5 }\).