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Either Let \( \omega = \cos \frac{\pi}{5} + i \sin \frac{\pi}{5} \) - CIE - A-Level Further Maths - Question 11 - 2011 - Paper 1
Question 11
Either Let \( \omega = \cos \frac{\pi}{5} + i \sin \frac{\pi}{5} \). Show that \( \omega^5 + 1 = 0 \) and deduce that \( \omega^4 + \omega^3 + \omega^2 - \omega = -... show full transcript
Worked Solution & Example Answer:Either Let \( \omega = \cos \frac{\pi}{5} + i \sin \frac{\pi}{5} \) - CIE - A-Level Further Maths - Question 11 - 2011 - Paper 1
Step 1
Show that \( \omega^5 + 1 = 0 \)
Answer
To show that ( \omega^5 + 1 = 0 ), we start with the definition of ( \omega ):
[ \omega = \cos \frac{\pi}{5} + i \sin \frac{\pi}{5} ] Using the properties of complex numbers, we can use De Moivre's theorem, which states that:
[ \omega^n = \cos(n\theta) + i\sin(n\theta) ] For ( n = 5 ) and ( \theta = \frac{\pi}{5} ), we find:
[ \omega^5 = \cos(\pi) + i\sin(\pi) = -1 + 0i ] Thus: [ \omega^5 + 1 = 0 ] This verifies that ( \omega^5 + 1 = 0 ).
Step 2
Show further that \( \omega - \omega^4 = 2 \cos \frac{\pi}{5} \)
Answer
First, calculate ( \omega^4 ):
Using De Moivre's theorem again: [ \omega^4 = \cos(4\times \frac{\pi}{5}) + i\sin(4\times \frac{\pi}{5}) = -\cos \frac{\pi}{5} + i\sin \frac{\pi}{5} ] Now, we subtract: [ \omega - \omega^4 = (\cos \frac{\pi}{5} + i\sin \frac{\pi}{5}) - (-\cos \frac{\pi}{5} + i\sin \frac{\pi}{5}) ] This simplifies to: [ \omega - \omega^4 = 2\cos \frac{\pi}{5} ]
Step 3
Show further that \( \omega^3 - \omega^2 = 2 \cos \frac{3\pi}{5} \)
Answer
We need to calculate ( \omega^3 ) and ( \omega^2 ):
Using De Moivre's theorem: [ \omega^3 = \cos(3\times \frac{\pi}{5}) + i\sin(3\times \frac{\pi}{5}) = -\cos \frac{2\pi}{5} + i\sin \frac{3\pi}{5} ] [ \omega^2 = \cos(2\times \frac{\pi}{5}) + i\sin(2\times \frac{\pi}{5}) ] Now perform the subtraction: [ \omega^3 - \omega^2 = (-\cos \frac{2\pi}{5} + i\sin \frac{3\pi}{5}) - (\cos \frac{2\pi}{5} + i\sin \frac{2\pi}{5}) ] This yields: [ \omega^3 - \omega^2 = 2\cos \frac{3\pi}{5} ]
Step 4
Find the values of \( \cos \frac{\pi}{5} + \cos \frac{3\pi}{5} \) and \( \cos \frac{\pi}{5} \)
Answer
From earlier equations, notice: [ \cos \frac{\pi}{5} + \cos \frac{3\pi}{5} = 1 ] To find ( \cos \frac{\pi}{5} ), we set: [ \cos \frac{\pi}{5} = x \ \cos \frac{3\pi}{5} = -1 - x ] Substituting into: [ 2\cos \frac{\pi}{5} = 1 \ \Rightarrow \cos \frac{\pi}{5} = \frac{1}{2} ]
Step 5
Find a quadratic equation having roots \( cos \frac{\pi}{5} \) and \( cos \frac{3\pi}{5} \)
Answer
Roots of the quadratic equation can be derived using the form: [ ax^2 + bx + c = 0 ] With the roots being ( r_1 = \cos \frac{\pi}{5} ) and ( r_2 = \cos \frac{3\pi}{5} ), we know: [ b = - (r_1 + r_2) \ c = r_1 \cdot r_2 ] Substituting: [ \Rightarrow x^2 - (\cos \frac{\pi}{5} + \cos \frac{3\pi}{5})x + \left( \cos \frac{\pi}{5} \cdot \cos \frac{3\pi}{5} \right) = 0 ] With the values calculated: This results in a specific form, which we can equate to find the necessary coefficients.