How do you multiply $(1 + 3 i) (1 - 3 i)$ $(1+3i)(1-3i)$ in trigonometric form?

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How do you multiply $(1 + 3 i) (1 - 3 i)$ $(1+3i)(1-3i)$ in trigonometric form?

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Answer 1 · 2016-03-27T23:56:13

$(1 + 3 i) (1 - 3 i) = 10$ $(1 + 3i)(1 - 3i) = 10$

Explanation:

$| 1 + 3 i | = | 1 - 3 i | = \sqrt{1^{2} + 3^{2}} = \sqrt{10}$ $abs(1+3i) = abs(1-3i) = sqrt(1^2+3^2) = sqrt(10)$

$1 + 3 i = \sqrt{10} c i s (arctan (3))$ $1 + 3i = sqrt(10) cis (arctan(3))$

$1 - 3 i = \sqrt{10} c i s (arctan (- 3)) = \sqrt{10} c i s (- arctan (3))$ $1 - 3i = sqrt(10) cis (arctan(-3)) = sqrt(10) cis (-arctan(3))$

So:

$(1 + 3 i) (1 - 3 i)$ $(1 + 3i)(1 - 3i)$

$= (\sqrt{10} c i s (arctan (3))) (\sqrt{10} c i s (- arctan (3)))$ $= (sqrt(10) cis (arctan(3)))(sqrt(10) cis (-arctan(3)))$

$= {(\sqrt{10})}^{2} c i s (arctan (3) - arctan (3))$ $= (sqrt(10))^2 cis (arctan(3) - arctan(3))$

$= 10 c i s (0)$ $= 10 cis(0)$

$= 10$ $= 10$

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How do you multiply $(1 + 3 i) (1 - 3 i)$ $(1+3i)(1-3i)$ in trigonometric form?

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How do you multiply (1+3i)(1-3i) (1+3i)(1−3i) (1+3i)(1-3i) in trigonometric form?

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How do you multiply $(1 + 3 i) (1 - 3 i)$ $(1+3i)(1-3i)$ in trigonometric form?