How do you add $(- 8 + 4 i)$ $(-8+4i)$ and $(- 3 - 3 i)$ $(-3-3i)$ in trigonometric form?

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How do you add $(- 8 + 4 i)$ $(-8+4i)$ and $(- 3 - 3 i)$ $(-3-3i)$ in trigonometric form?

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Answer 1 · 2016-02-15T09:17:45

$\sqrt{122} [cos ({tan}^{- 1} (\frac{1}{- 11})) + i sin ({tan}^{- 1} (\frac{1}{- 11}))]$ $sqrt(122)[cos(tan^-1(1/(-11)))+isin(tan^-1(1/(-11)))]" " "$ OR
$\sqrt{122} [cos ({174.806}^{\circ}) + i sin ({174.806}^{\circ})]$ $sqrt(122)[cos(174.806^@)+isin(174.806^@)]" " "$

Explanation:

Add $(- 8 + 4 i)$ $(-8+4i)$ and $(- 3 - 3 i)$ $(-3-3i)$ to obtain

$- 11 + i$ $-11+i$

Use the formula

$a + b i = \sqrt{a^{2} + b^{2}} [cos ({tan}^{- 1} (\frac{b}{a})) + i \cdot sin ({tan}^{- 1} (\frac{b}{a})]$ $a+bi=sqrt(a^2+b^2)[cos(tan^-1 (b/a))+i*sin(tan^-1 (b/a)]$

Let $a = - 11$ $a=-11$ and $b = 1$ $b=1$

magnitude $r = \sqrt{a^{2} + b^{2}} = \sqrt{{(- 11)}^{2} + 1^{2}} = \sqrt{122}$ $" " " "r=sqrt(a^2+b^2)=sqrt((-11)^2+1^2)=sqrt122$

the angle $θ = {tan}^{- 1} (\frac{1}{- 11}) = {174.806}^{\circ}$ $theta=tan^-1 (1/(-11))=174.806^@$

$\sqrt{122} [cos ({tan}^{- 1} (\frac{1}{- 11})) + i sin ({tan}^{- 1} (\frac{1}{- 11}))]$ $sqrt(122)[cos(tan^-1(1/(-11)))+isin(tan^-1(1/(-11)))]" " "$ OR
$\sqrt{122} [cos ({174.806}^{\circ}) + i sin ({174.806}^{\circ})]$ $sqrt(122)[cos(174.806^@)+isin(174.806^@)]" " "$

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How do you add $(- 8 + 4 i)$ $(-8+4i)$ and $(- 3 - 3 i)$ $(-3-3i)$ in trigonometric form?

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Explanation:

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How do you add (-8+4i)(−8+4i)(-8+4i) and (-3-3i)(−3−3i)(-3-3i) in trigonometric form?

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How do you add $(- 8 + 4 i)$ $(-8+4i)$ and $(- 3 - 3 i)$ $(-3-3i)$ in trigonometric form?