Question #bee24

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Question #bee24

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Answer 1 · 2017-11-18T16:20:05

$12 i - 5$ $12i - 5$

Explanation:

${(2 + 3 i)}^{2}$ $(2 + 3i)^2$

Expand first and treat $i$ $i$ as if it were a variable
${(2 + 3 i)}^{2}$ $(2 + 3i)^2$
$4 + 12 i + 9 i^{2}$ $4 + 12i + 9i^2$

Since we know that
$i^{1} = i$ $i^1 = i$
$i^{2} = - 1$ $i^2 = -1$
$i^{3} = - i$ $i^3 = -i$
$i^{4} = 1$ $i^4 = 1$
We can substitute the value of $i^{2}$ $i^2$ with $- 1$ $-1$

$4 + 12 i + 9 i^{2}$ $4 + 12i + 9i^2$
$4 + 12 i + 9 (- 1)$ $4 + 12i + 9(-1)$
$4 + 12 i - 9$ $4 + 12i - 9$
$12 i - 5$ $12i - 5$

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Question #bee24

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