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

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

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

simplify

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

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

Solution steps

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

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

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

(e - 23) (e - 23)

(4 - 4 sin (x))^{2}

3 (x - 4) - 4 (2 x - d = 5)

(1 + 2 x) d x

(x - 1) (x - (2 + i)) (x - (2 - i))

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