How do you add $(- 6 + i) + (9 + 3 i)$ $(-6+i)+(9+3i)$ in trigonometric form?

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How do you add $(- 6 + i) + (9 + 3 i)$ $(-6+i)+(9+3i)$ in trigonometric form?

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Answer 1 · 2018-08-06T04:15:20

$(- 6 + i) + (9 + 3 i) = 3 + 4 i, I Quadrant$ $color(maroon)((-6 + i) + (9 + 3i) = 3 + 4 i, " I Quadrant"$

Explanation:

$(- 6 + i) + (9 + 3 i)$ $(-6 + i) + (9 + 3i)$

$z = x + i y$ $z = x + i y$

$z = r (cos θ + i sin θ)$ $z = r (cos theta + i sin theta)$

$r = | z | = ∣ ∣ ∣ \sqrt{x^{2} + y^{2}} ∣ ∣ ∣$ $r = |z| = |sqrt(x^2 + y^2)|$

$r_{1} = \sqrt{- 6^{2} + 1^{2}} = \sqrt{37}$ $r_1 = sqrt(-6^2 + 1^2) = sqrt37$

$θ_{1} = arctan (\frac{y}{x}) = {tan}^{- 1} (\frac{1}{- 6}) = {170.5377}^{\circ}, II Quadrant$ $theta_1 = arctan (y/x) = tan^-1 (1/-6) = 170.5377^@, " II Quadrant"$

$r_{2} = \sqrt{9^{2} + 3^{2}} = \sqrt{90}$ $r_2 = sqrt(9^2 + 3^2) = sqrt90$

$θ_{2} = arctan (\frac{y}{x}) = {tan}^{- 1} (\frac{3}{9}) = {18.4349}^{\circ}, I Quadrant$ $theta_2 = arctan (y/x) = tan^-1 (3/9) = 18.4349^@, " I Quadrant"$

$(- 6 + i) + (9 + 3 i) = \sqrt{37} (cos 170.5377 + sin 170.5377) + \sqrt{90} (cos 18.4349 + sin 18.4349)$ $(-6 + i) + (9 + 3i) = sqrt37(cos 170.5377 + sin170.5377) + sqrt90 (cos 18.4349 + sin 18.4349)$

$\Rightarrow \sqrt{37} cos 170.5377 + \sqrt{90} cos 18.4349 + i (\sqrt{37} sin 170.5377 + \sqrt{90} sin 18.4349)$ $=> sqrt37 cos 170.5377 + sqrt90 cos 18.4349 + i (sqrt37 sin 170.5377 + sqrt90 sin 18.4349)$

$\Rightarrow - 6 + 9 + 1 i + 3 i$ $=> -6 + 9 + 1 i + 3 i$

$(- 6 + i) + (9 + 3 i) = 3 + 4 i, I Quadrant$ $color(maroon)((-6 + i) + (9 + 3i) = 3 + 4 i, " I Quadrant"$

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How do you add $(- 6 + i) + (9 + 3 i)$ $(-6+i)+(9+3i)$ in trigonometric form?

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How do you add (−6+i)+(9+3i)(-6+i)+(9+3i) in trigonometric form?

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Explanation:

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How do you add $(- 6 + i) + (9 + 3 i)$ $(-6+i)+(9+3i)$ in trigonometric form?