First slide

De-moivre's theorem

Question

$If α is any of 7^{th} roots of unity, then α = \frac{cis 2 K π}{7} (K = 0 to 6) and \sum_{i = 1}^{6} α^{i} = - 1 (α \neq 1) and α^{7} = 1 . Then answer the following$

Difficult

Question

$The equation whose roots are α + α^{2} + α^{4} and α^{3} + α^{5} + α^{6} is (α \neq 1)$

A

$x^{2} + x - 2 = 0$
B

$x^{2} + x + 2 = 0$
C

$x^{2} - x + 2 = 0$
D

$x^{2} - x - 2 = 0$

Solution

$\begin{array}{l} Let α = cis \frac{2 π}{7} then a = α + α^{2} + α^{4}, b = α^{3} + α^{5} + α^{6} \\ \Rightarrow a + b = α + α^{2} + α^{3} + α^{4} + α^{5} + α^{6} = - 1 & a b = (α + α^{2} + α^{4}) (α^{3} + α^{5} + α^{6}) = 2 \end{array}$
$∴ Equation whose roots are a, b is x^{2} - x (- 1) + 2 = 0$

Question

$If f (x) = 1 + 2 x + 3 x^{2} + 4 x^{3} + 5 x^{4} + 6 x^{5} + 7 x^{6} then for α \neq 1,$ $f (x) + f (α x) + f (α^{2} x) + f (α^{3} x) + - - - + f (α^{6} x)$ =

A

42
B

21
C

14
D

7

Solution

$\begin{array}{l} f (x) + f (α x) + f (α^{2} x) + f (α^{3} x) + - - - + f (α^{6} x) \\ = (1 + 1 + 1 - - + 1) + 2 x (1 + α + α^{2} + α^{3} + - - - + α^{6}) + 3 x^{2} (1 + α^{2} + α^{4} + - - - + α^{12}) + - - - - + \\ = 7 x^{6} (1 + α^{6} + α^{12} + α^{18} + - - - + α^{36}) = 7 \end{array}$

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