What is e^{(-2+ipin)x} ?

The solution to e^{(-2+ipin)x} is e^{-2x}cos(pinx)+ie^{-2x}sin(pinx)

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simplify e^{(-2+ipin)x}

simplify

e^{(- 2 + iπn) x}

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

Solution steps

e^{(- 2 + iπn) x}

Apply imaginary number rule:

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

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

Rewrite

e^{- 2 x} (cos (πn x) + sin (πn x) i)

in standard complex form:

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

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

1 + (3 x^{2})^{4}

\frac{3 x + 79}{2 ( x - 3 ) ( x + 8 )}

5 (2 t - 3) - 3 (2 t - 1 = 0)

A (4, - 3, - 2)

\frac{d p}{p ( 1 - p )}

What is e^{(-2+ipin)x} ?
The solution to e^{(-2+ipin)x} is e^{-2x}cos(pinx)+ie^{-2x}sin(pinx)