What is (e^{iθ}-1)(e^{-iθ}+1) ?

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

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simplify (e^{iθ}-1)(e^{-iθ}+1)

simplify

(e^{i θ} - 1) (e^{- i θ} + 1)

2 i sin (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) (c + d i) = (a c - b d) + (a d + b c) i

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

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

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

Simplify

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

= 0 + 2 sin (θ) i

Refine

= 2 sin (θ) i

\frac{3 x + 5}{2 ( x + 2 )}

3 (2 cos (x) - x sin (x))

x \cdot (x + 1)

(2 sin (t) + 2)^{2}

(- 3 - i)^{5}

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