How do you add $(- 9 + i) + (- 1 + 3 i)$ $(-9+i)+(-1+3i)$ in trigonometric form?

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How do you add $(- 9 + i) + (- 1 + 3 i)$ $(-9+i)+(-1+3i)$ in trigonometric form?

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Answer 1 · 2018-07-09T11:33:33

$\Rightarrow - 10 + 4 i$ $color(crimson)(=> -10+ 4i$

Explanation:

$z = a + b i = r (cos θ + i sin θ)$ $z= a+bi= r (costheta+isintheta)$

$r = \sqrt{a^{2} + b^{2}}, θ = {tan}^{- 1} (\frac{b}{a})$ $r=sqrt(a^2+b^2), " " theta=tan^-1(b/a)$

$r_{1} (cos (θ_{1}) + i sin (θ_{2})) + r_{2} (cos (θ_{2}) + i sin (θ_{2})) = r_{1} cos (θ_{1}) + r_{2} cos (θ_{2}) + i (r_{1} sin (θ_{1}) + r_{2} sin (θ_{2}))$ $r_1(cos(theta_1)+isin(theta_2))+r_2(cos(theta_2)+isin(theta_2))=r_1cos(theta_1)+r_2cos(theta_2)+i(r_1sin(theta_1)+r_2sin(theta_2))$

$r_{1} = \sqrt{- 9^{2} + 1^{2}}) = \sqrt{82}$ $r_1=sqrt(-9^2+ 1^2))=sqrt 82$
$r_{2} = \sqrt{- 1^{2} + 3^{2}} = \sqrt{10}$ $r_2=sqrt(-1^2+ 3^2) =sqrt 10$

$θ_{1} = {tan}^{- 1} (\frac{1}{- 9}) \approx {173.66}^{\circ}, II quadrant$ $theta_1=tan^-1(1 / -9)~~ 173.66^@, " II quadrant"$
$θ_{2} = {tan}^{- 1} (\frac{3}{- 1}) \approx {108.43}^{\circ}, II quadrant$ $theta_2=tan^-1(3/ -1)~~ 108.43^@, " II quadrant"$

$z_{1} + z_{2} = \sqrt{82} cos (173.66) + \sqrt{10} cos (108.43) + i (\sqrt{82} sin 173.66 + \sqrt{10} sin 108.43)$ $z_1 + z_2 = sqrt 82 cos(173.66) + sqrt 10 cos(108.43) + i (sqrt 82 sin 173.66 + sqrt 10 sin 108.43)$

$\Rightarrow - 9 - 1 + i (1 + 3)$ $=> -9 - 1 + i (1 + 3 )$

$\Rightarrow - 10 + 4 i$ $color(crimson)(=> -10+ 4i$

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How do you add $(- 9 + i) + (- 1 + 3 i)$ $(-9+i)+(-1+3i)$ in trigonometric form?

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How do you add (−9+i)+(−1+3i)(-9+i)+(-1+3i) in trigonometric form?

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How do you add $(- 9 + i) + (- 1 + 3 i)$ $(-9+i)+(-1+3i)$ in trigonometric form?