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

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

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

simplify

e^{(- 1 - i) t}

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

Solution steps

e^{(- 1 - i) t}

Apply imaginary number rule:

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

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

cos (- t) = cos (t)

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

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

Rewrite

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

in standard complex form:

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

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

(3 + x 3)^{2}

\frac{1}{3} (1 = 3 A + B + (- 9) C + D) \frac{1}{3}

(3 + 2 z i)^{2}

\frac{1}{81} (sin (9 (t - x))) + 9 x cos ((t - x) 9)

- \frac{n ( d ^{2} v + e d v + d v f - d b + e v f - e b )}{m c ( d + e )}