What is e^{(-2+3i)t} ?

The solution to e^{(-2+3i)t} is e^{-2t}cos(3t)+ie^{-2t}sin(3t)

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simplify e^{(-2+3i)t}

simplify

e^{(- 2 + 3 i) t}

e^{- 2 t} cos (3 t) + i e^{- 2 t} sin (3 t)

Solution steps

e^{(- 2 + 3 i) t}

Apply imaginary number rule:

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

= e^{- 2 t} (cos (3 t) + i sin (3 t))

Rewrite

e^{- 2 t} (cos (3 t) + sin (3 t) i)

in standard complex form:

e^{- 2 t} cos (3 t) + e^{- 2 t} sin (3 t) i

= e^{- 2 t} cos (3 t) + e^{- 2 t} sin (3 t) i

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

\frac{2 ( 6 - x ^{2} ) ^{2}}{x ^{2}}

2 π - 2 π x (2 - 2 x^{2})

What is e^{(-2+3i)t} ?
The solution to e^{(-2+3i)t} is e^{-2t}cos(3t)+ie^{-2t}sin(3t)