化简 (1-e^{ix})(1-e^{-ix})

解答

化简

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

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

求解步骤

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

使用虚数运算法则:

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)

使用复数算数法则:

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

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

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

化简

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

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

e x p an d \frac{36 ( - e ^{5 π} + 1 )}{725 e ^{5 π}}

e x p an d \frac{2 \cdot ( n + 1 + n \cdot 3 ^{n} + 3 ^{n} )}{n + 3 ^{n + 1}}

e x p an d (x^{2} - 5) e^{x}

e x p an d \frac{( 2 x + 7 ) ^{2}}{4}

e x p an d 0.03 \cdot x (1. 2^{x})