a-b-are-complex-numbers-and-a-2-ab-b-2-0-find-a-a-b-2020-b-a-b-2020-

Question Number 88314 by M±th+et£s last updated on 09/Apr/20

(a, b) a r e c o m p l e x n u m b e r s a n d a^{2} + a b + b^{2} = 0

f i n d {(\frac{a}{a + b})}^{2020} + {(\frac{b}{a + b})}^{2020}

Commented by mathmax by abdo last updated on 09/Apr/20

a i s s l u t i o n o f x^{2} + b x + b^{2} = 0

Δ = b^{2} - 4 b^{2} = - 3 b^{2} = {(i \sqrt{3} b)}^{2} \Rightarrow x_{1} = \frac{- b + i \sqrt{3} b}{2} = b e^{\frac{i 2 π}{3}}

a n d x_{2} = \frac{- b - i \sqrt{b}}{2} = {b e}^{- \frac{i 2 π}{3}}

a = x_{1} \Rightarrow \frac{a}{a + b} = \frac{b e^{\frac{i 2 π}{3}}}{{b e}^{\frac{i 2 π}{3}} + b} = \frac{e^{\frac{i 2 π}{3}}}{1 + e^{\frac{i 2 π}{3}}} = \frac{1}{1 + e^{- \frac{i 2 π}{3}}} = \frac{1}{1 - \frac{1}{2} - \frac{i \sqrt{3}}{2}} = = \frac{1}{\frac{1}{2} - \frac{i \sqrt{3}}{2}} = e^{\frac{i π}{3}}

\frac{b}{a + b} = \frac{b}{{b e}^{\frac{i 2 π}{3}} + b} = \frac{1}{e^{\frac{i 2 π}{3}} + 1} = \frac{1}{1 - \frac{1}{2} + \frac{i \sqrt{3}}{2}} = \frac{1}{\frac{1}{2} + \frac{i \sqrt{3}}{2}} = e^{- \frac{i π}{3}} \Rightarrow

{(\frac{a}{a + b})}^{2020} + {(\frac{b}{a + b})}^{2020} = {(e^{\frac{i π}{3}})}^{2020} + {(e^{- \frac{i π}{3}})}^{2020}

= 2 c o s (\frac{2020 π}{3}) t h i s v a l u e c a n b e s i m p l i f i e d \dots

w r f i n d t h e s a m e v a l u e f o r x = x_{2}

Answered by MJS last updated on 09/Apr/20

a^{2} + a b + b^{2} = 0

\Rightarrow b = a (- \frac{1}{2} \pm \frac{\sqrt{3}}{2} i)

\Rightarrow \frac{a}{a + b} = \frac{1}{2} \mp \frac{\sqrt{3}}{2} i \land \frac{b}{a + b} = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i

\Rightarrow {(\frac{a}{a + b})}^{n} + {(\frac{b}{a + b})}^{n} = e^{- i \frac{n π}{3}} + e^{i \frac{n π}{3}}

now you should be able to find the answer

Commented by M±th+et£s last updated on 09/Apr/20

t h a n x s i r . i t h i n k a n s = - 1

Commented by MJS last updated on 09/Apr/20

yes

Answered by Joel578 last updated on 10/Apr/20

{(a + b)}^{2} - a b = 0 \Rightarrow {(a + b)}^{2} = a b \Rightarrow \frac{a b}{{(a + b)}^{2}} = 1

\Rightarrow (\frac{a}{a + b}) (\frac{b}{a + b}) = 1

(\frac{a}{a + b}) + (\frac{b}{a + b}) = \frac{a + b}{a + b} = 1

Let x = \frac{a}{a + b}, y = \frac{b}{a + b} with x y = 1, x + y = 1

Define f (n) = x^{n} + y^{n}

f (1) = x + y = 1

f (2) = x^{2} + y^{2} = {(x + y)}^{2} - 2 x y = - 1

f (3) = x^{3} + y^{3} = {(x + y)}^{3} - 3 x y (x + y) = - 2

f (4) = x^{4} + y^{4} = (x^{3} + y^{3}) (x + y) - x y (x^{2} + y^{2}) = - 1

f (5) = 1, f (6) = 2, f (7) = 1, f (8) = - 1, f (9) = - 2

f (n) = f (n + 6)

∴ f (2020) = f (4) = - 1

Commented by Joel578 last updated on 10/Apr/20

without complex method

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