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

The solution to i(1+2e^{it}) is -2sin(t)+i(1+2cos(t))

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

simplify

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

- 2 sin (t) + i (1 + 2 cos (t))

Solution steps

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

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

= i (2 (cos (t) + i sin (t)) + 1)

Apply complex arithmetic rule:

(ai) (b + c i) = - a c + abi

a = 1, b = 1 + 2 cos (t), c = 2 sin (t)

= - 1 \cdot 2 sin (t) + 1 \cdot (1 + 2 cos (t)) i

Simplify

- 1 \cdot 2 sin (t) : - 2 sin (t)

Simplify

1 \cdot (1 + 2 cos (t)) : 1 + 2 cos (t)

= - 2 sin (t) + (1 + 2 cos (t)) i

\frac{( - 2 a + 7 b ) ^{2}}{9}

- \frac{18 x}{( 3 x ^{2} + 1 ) ^{2}}

\frac{6 x - 5}{3 ( x - 1 ) ^{\frac{2}{3}} ( 2 x - 1 ) ^{\frac{1}{3}}}

(x^{3} + \frac{1}{2 x})^{15}