What is (e^{iθ}-1)(e^{-iθ}-1) ?

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

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simplify (e^{iθ}-1)(e^{-iθ}-1)

simplify

(e^{i θ} - 1) (e^{- i θ} - 1)

(cos (θ) - 1)^{2} + sin^{2} (θ)

Solution steps

(e^{i θ} - 1) (e^{- i θ} - 1)

Apply imaginary number rule:

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

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

e^{- i θ} = cos (θ) - i sin (θ)

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

Apply complex arithmetic rule:

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

a = cos (θ) - 1, b = sin (θ)

= (cos (θ) - 1)^{2} + sin^{2} (θ)

2 (x + \frac{1}{x}) (1 - \frac{1}{x ^{2}})

lo g_{10} (x^{3} y^{- 12})

(x - 1) (x + \frac{1}{2})

e^{t} (3 cos (5 t) - 2 sin (5 s))

\frac{10 s}{( s + 2 ) ^{3} ( s + 8 )}

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