What is (1-e^{ix})(1-e^{-ix}) ?

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

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simplify (1-e^{ix})(1-e^{-ix})

simplify

(1 - e^{i x}) (1 - e^{- i x})

(1 - cos (x))^{2} + sin^{2} (x)

Solution steps

(1 - e^{i x}) (1 - e^{- i x})

Apply imaginary number rule:

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

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

e^{- i x} = cos (x) - i sin (x)

= (- (cos (x) + i sin (x)) + 1) (- (cos (x) - i sin (x)) + 1)

Apply complex arithmetic rule:

(a + bi) (a - bi) = a^{2} + b^{2}

a = 1 - cos (x), b = - sin (x)

= (1 - cos (x))^{2} + (- sin (x))^{2}

Simplify

(1 - cos (x))^{2} + (- sin (x))^{2} : (1 - cos (x))^{2} + sin^{2} (x)

= (1 - cos (x))^{2} + sin^{2} (x)

\frac{36 ( - e ^{5 π} + 1 )}{725 e ^{5 π}}

\frac{2 \cdot ( n + 1 + n \cdot 3 ^{n} + 3 ^{n} )}{n + 3 ^{n + 1}}

(x^{2} - 5) e^{x}

\frac{( 2 x + 7 ) ^{2}}{4}

0.03 \cdot x (1. 2^{x})

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