What is e^{(1+2i)x} ?

The solution to e^{(1+2i)x} is e^xcos(2x)+ie^xsin(2x)

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simplify e^{(1+2i)x}

simplify

e^{(1 + 2 i) x}

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

Solution steps

e^{(1 + 2 i) x}

Apply imaginary number rule:

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

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

Rewrite

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

in standard complex form:

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

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

(1 - t) (4, 6, 6) + t (6, 7, 2)

(3 x - 5 \circ - 4 x - 2) (- 5)

(x + y - 1) d x

(u^{9} + u) (cos (x))

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