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

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

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simplify ie^{-(1+i)t}

simplify

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

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

Solution steps

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

e^{- (1 + i) t} = e^{- t} (cos (t) - i sin (t))

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

Apply complex arithmetic rule:

(ai) (b + c i) = - a c + abi

a = e^{- t}, b = cos (t), c = - sin (t)

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

Simplify

- e^{- t} (- sin (t)) : e^{- t} sin (t)

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

(x - e^{x} sin (y)) d y

x - 1 (1 - 6) - 6

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

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

(2 x y - y) d x + x d y

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