What is e^{2iθ}(cos(2θ)-isin(2θ)) ?

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

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simplify e^{2iθ}(cos(2θ)-isin(2θ))

simplify

e^{2 i θ} (cos (2 θ) - i sin (2 θ))

1

Solution steps

e^{2 i θ} (cos (2 θ) - i sin (2 θ))

e^{2 i θ} = cos (2 θ) + i sin (2 θ)

= (cos (2 θ) + i sin (2 θ)) (cos (2 θ) - i sin (2 θ))

Apply complex arithmetic rule:

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

a = cos (2 θ), b = sin (2 θ)

= cos^{2} (2 θ) + sin^{2} (2 θ)

cos^{2} (2 θ) + sin^{2} (2 θ) = 1

= 1

\frac{x - 6 y}{5 ( x - 2 y )}

\frac{d}{d θ} ((r (θ)) (1 - 2 θ))

\frac{( x - 2 ) ^{4}}{( x - 3 ) ^{5}}

(x ln (x) - 1) (x ln (x) + 1)

(x^{2} - x) (e^{3 x})

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