How do you add $(- 5 - 2 i) + (7 - 6 i)$ $(-5-2i)+(7-6i)$ in trigonometric form?

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How do you add $(- 5 - 2 i) + (7 - 6 i)$ $(-5-2i)+(7-6i)$ in trigonometric form?

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Answer 1 · 2017-07-15T17:31:01

$(- 5 - 2 i) + (7 - 6 i) = 2 \sqrt{17} (cos 1.33 - i sin 1.33)$ $(-5-2i) +(7-6i) = 2sqrt17(cos1.33-isin1.33)$

Explanation:

$(- 5 - 2 i) + (7 - 6 i) = 2 - 8 i$ $(-5-2i)+(7-6i) = 2-8i$

This can be written in trigonometric form as:

$a + b i = r (cos ϑ + i sin ϑ)$ $a+bi = r(cosvartheta + i sinvartheta)$ where $r = \sqrt{a^{2} + b^{2}}$ $r=sqrt (a^2+b^2)$ and $ϑ = arctan (\frac{y}{x})$ $vartheta = arctan(y/x)$

$\sqrt{2^{2} + {(- 8)}^{2}} = 2 \sqrt{17}$ $sqrt(2^2+(-8)^2) = 2sqrt17$

$arctan (- \frac{8}{2}) = - 1.33$ $arctan(-8/2) = -1.33$

$therefore 2-8i = 2sqrt17(cos(-1.33) + isin(-1.33))$

Using the properties of the $sin$ and $cos$ functions, this can be rewritten as:

$2-8i=2sqrt17(cos1.33-isin1.33)$

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How do you add $(- 5 - 2 i) + (7 - 6 i)$ $(-5-2i)+(7-6i)$ in trigonometric form?

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Explanation:

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How do you add $(- 5 - 2 i) + (7 - 6 i)$ $(-5-2i)+(7-6i)$ in trigonometric form?