If $i^{2} = - 1$ , calculate the value of $3 i^{2} + i^{3} - i^{4}$ .

- 4 - i

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- 2 - i

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2 + i

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4 + i

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6 + 2 i

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Solution

The correct option is A $- 4 - i$
$i$ is an imaginary number whose value is $\sqrt{- 1}$
So, $i^{2} = - 1$
$i^{3} = i^{2} * i = - 1 * i = - i$
$i^{4} = (i^{2})^{2} = {(- 1)}^{2} = 1$
So the value of $3 i^{2} + i^{3} - i^{4}$ is
$\Rightarrow 3 \times (- 1) + (- i) - (1)$
$\Rightarrow - 3 - i - 1 = - 4 - i$

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Similar questions

(1 + i + i^{2} + i^{3} + i^{4} + i^{5}) (1 + i) =

i = \sqrt{- 1}

i^{2} = - 1

i^{3} = - i

i^{4} = 1

i^{127}

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a + i b

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z {(2 - 2 \sqrt{3} i)}^{2} = i {(\sqrt{3} + i)}^{4}

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