What is e^{(2+3i)t} ?

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

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simplify e^{(2+3i)t}

simplify

e^{(2 + 3 i) t}

e^{2 t} cos (3 t) + i e^{2 t} sin (3 t)

Solution steps

e^{(2 + 3 i) t}

Apply imaginary number rule:

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

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

Rewrite

e^{2 t} (cos (3 t) + sin (3 t) i)

in standard complex form:

e^{2 t} cos (3 t) + e^{2 t} sin (3 t) i

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

(1 - 2 x^{2}) y^{''} (x) - 10 x y^{'} (x) - 6 y

\frac{d}{d t} ((x) (- 2611 t^{2} + 29112 t + 6844))

A (- 3, - 2) B (1, - 2) C (1, 2)

4 a (a + a)

(x^{3} - 3 x y)^{2} + (3 x^{2} y - y^{3})^{2}

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