What is e^{(2-3i)x} ?

The solution to e^{(2-3i)x} is e^{2x}cos(3x)-ie^{2x}sin(3x)

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simplify e^{(2-3i)x}

simplify

e^{(2 - 3 i) x}

e^{2 x} cos (3 x) - i e^{2 x} sin (3 x)

Solution steps

e^{(2 - 3 i) x}

Apply imaginary number rule:

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

= e^{2 x} (cos (- 3 x) + i sin (- 3 x))

cos (- 3 x) = cos (3 x)

sin (- 3 x) i = - i sin (3 x)

= e^{2 x} (cos (3 x) - i sin (3 x))

Rewrite

e^{2 x} (cos (3 x) - i sin (3 x))

in standard complex form:

e^{2 x} cos (3 x) - e^{2 x} sin (3 x) i

= e^{2 x} cos (3 x) - e^{2 x} sin (3 x) i

(0.4 e^{6 T} - 0.4 e^{T}) 2 z

3 (\frac{x}{3} - 10 = - 7)

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

\frac{d}{d t} ((x) (t^{3} + t - 3))

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

What is e^{(2-3i)x} ?
The solution to e^{(2-3i)x} is e^{2x}cos(3x)-ie^{2x}sin(3x)