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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

simplify

e^{(- 2 + i) t}

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

Solution steps

e^{(- 2 + i) t}

Apply imaginary number rule:

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

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

Rewrite

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

in standard complex form:

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

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

A (- 5, - 3) P (- 9, 5)

(e^{x} - e^{- x}) (e^{x})

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

f (x + d x, y, z) - f (x, y, z)

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

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