What is (1+i)e^{-3it} ?

The solution to (1+i)e^{-3it} is (cos(3t)+sin(3t))+i(cos(3t)-sin(3t))

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simplify (1+i)e^{-3it}

simplify

(1 + i) e^{- 3 i t}

(cos (3 t) + sin (3 t)) + i (cos (3 t) - sin (3 t))

Solution steps

(1 + i) e^{- 3 i t}

e^{- 3 i t} = cos (3 t) - i sin (3 t)

= (1 + i) (cos (3 t) - i sin (3 t))

Apply complex arithmetic rule:

(a + bi) (c + d i) = (a c - b d) + (a d + b c) i

a = 1, b = 1, c = cos (3 t), d = - sin (3 t)

= (1 \cdot cos (3 t) - 1 \cdot (- sin (3 t))) + (1 \cdot (- sin (3 t)) + 1 \cdot cos (3 t)) i

Simplify

1 \cdot cos (3 t) - 1 \cdot (- sin (3 t)) : cos (3 t) + sin (3 t)

Simplify

1 \cdot (- sin (3 t)) + 1 \cdot cos (3 t) : - sin (3 t) + cos (3 t)

= (cos (3 t) + sin (3 t)) + (- sin (3 t) + cos (3 t)) i

Remove parentheses:

(a) = a

= cos (3 t) + sin (3 t) + (- sin (3 t) + cos (3 t)) i

Rewrite

cos (3 t) + sin (3 t) + (- sin (3 t) + cos (3 t)) i

in standard complex form:

(cos (3 t) + sin (3 t)) + (- sin (3 t) + cos (3 t)) i

= (cos (3 t) + sin (3 t)) + (- sin (3 t) + cos (3 t)) i

\frac{10}{s ( s ^{2} + 7 s + 10 )}

(x + y) d x + x^{2} + y^{2} d y

x (2 x - 1)^{2} e^{x}

\frac{( ( n + 1 ) ^{4} )}{n}

\frac{( x + 1 ) ( x + 2 )}{( 2 x + 1 ) ( x - 3 )}

What is (1+i)e^{-3it} ?
The solution to (1+i)e^{-3it} is (cos(3t)+sin(3t))+i(cos(3t)-sin(3t))