What is e^{(-2-2i)t} ?

The solution to e^{(-2-2i)t} is e^{-2t}cos(2t)-ie^{-2t}sin(2t)

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simplify e^{(-2-2i)t}

simplify

e^{(- 2 - 2 i) t}

e^{- 2 t} cos (2 t) - i e^{- 2 t} sin (2 t)

Solution steps

e^{(- 2 - 2 i) t}

Apply imaginary number rule:

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

= e^{- 2 t} (cos (- 2 t) + i sin (- 2 t))

cos (- 2 t) = cos (2 t)

sin (- 2 t) i = - i sin (2 t)

= e^{- 2 t} (cos (2 t) - i sin (2 t))

Rewrite

e^{- 2 t} (cos (2 t) - i sin (2 t))

in standard complex form:

e^{- 2 t} cos (2 t) - e^{- 2 t} sin (2 t) i

= e^{- 2 t} cos (2 t) - e^{- 2 t} sin (2 t) i

(ln (y) - 1)^{2}

\frac{76 a ^{2} b ^{3}}{19 ab}

\frac{s x + d}{( x - c ) ^{2}}

ln (x^{3} + 2) (x^{2} + 3)

What is e^{(-2-2i)t} ?
The solution to e^{(-2-2i)t} is e^{-2t}cos(2t)-ie^{-2t}sin(2t)