What is e^{-iθ}(1+e^{iθ}+e^{2iθ}) ?

The solution to e^{-iθ}(1+e^{iθ}+e^{2iθ}) is cos(θ)(1+cos(θ)+cos(2θ))+sin(θ)(sin(θ)+sin(2θ))

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simplify e^{-iθ}(1+e^{iθ}+e^{2iθ})

simplify

e^{- i θ} (1 + e^{i θ} + e^{2 i θ})

cos (θ) (1 + cos (θ) + cos (2 θ)) + sin (θ) (sin (θ) + sin (2 θ))

Solution steps

e^{- i θ} (1 + e^{i θ} + e^{2 i θ})

e^{- i θ} = cos (θ) - i sin (θ)

= (cos (θ) - i sin (θ)) (e^{i θ} + e^{2 i θ} + 1)

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= (cos (θ) - i sin (θ)) (e^{2 i θ} + cos (θ) + i sin (θ) + 1)

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= (cos (θ) - i sin (θ)) (cos (θ) + i sin (θ) + cos (2 θ) + i sin (2 θ) + 1)

Apply complex arithmetic rule:

(a + bi) (c + d i) = (a c - b d) + (a d + b c) i

a = cos (θ), b = - sin (θ), c = 1 + cos (θ) + cos (2 θ), d = sin (θ) + sin (2 θ)

= (cos (θ) (1 + cos (θ) + cos (2 θ)) - (- sin (θ)) (sin (θ) + sin (2 θ))) + (cos (θ) (sin (θ) + sin (2 θ)) + (- sin (θ)) (1 + cos (θ) + cos (2 θ))) i

Simplify

cos (θ) (1 + cos (θ) + cos (2 θ)) - (- sin (θ)) (sin (θ) + sin (2 θ)) : cos (θ) (1 + cos (θ) + cos (2 θ)) + sin (θ) (sin (θ) + sin (2 θ))

cos (θ) (sin (θ) + sin (2 θ)) + (- sin (θ)) (1 + cos (θ) + cos (2 θ)) = 0

= (cos (θ) (1 + cos (θ) + cos (2 θ)) + sin (θ) (sin (θ) + sin (2 θ))) + 0 i

Refine

= cos (θ) (1 + cos (θ) + cos (2 θ)) + sin (θ) (sin (θ) + sin (2 θ))

\frac{2 t + 1}{2 ( t ^{2} + t )}

a (- 5, - 1) b (- 1, 1)

\frac{A K N ^{2}}{x ( x ^{2} + N ^{2} )}

\frac{f}{( 2 - 2 f ) ^{2} ( 3 - f )}

\frac{6}{s ^{4} ( s ^{2} + 1 )}

What is e^{-iθ}(1+e^{iθ}+e^{2iθ}) ?
The solution to e^{-iθ}(1+e^{iθ}+e^{2iθ}) is cos(θ)(1+cos(θ)+cos(2θ))+sin(θ)(sin(θ)+sin(2θ))