What is e^{(-1+2i)t} ?

The solution to e^{(-1+2i)t} is e^{-t}cos(2t)+ie^{-t}sin(2t)

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simplify e^{(-1+2i)t}

simplify

e^{(- 1 + 2 i) t}

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

Solution steps

e^{(- 1 + 2 i) t}

Apply imaginary number rule:

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

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

Rewrite

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

in standard complex form:

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

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

2.7 (5.1 x + 4.9 = 3.2 x + 28.9)

- \frac{5 x - 1}{x ( x - 1 ) ^{2} ( 2 x + 3 )}

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

A (- 2, - 4) y B (5, 4)

(1 + y) d y

What is e^{(-1+2i)t} ?
The solution to e^{(-1+2i)t} is e^{-t}cos(2t)+ie^{-t}sin(2t)