How do you add $(8 - 9 i) + (4 + 6 i)$ $(8-9i)+(4+6i)$ in trigonometric form?

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How do you add $(8 - 9 i) + (4 + 6 i)$ $(8-9i)+(4+6i)$ in trigonometric form?

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Answer 1 · 2018-06-25T07:57:38

$\Rightarrow 12 - 3 i$ $color(gray)(=> 12 - 3 i$

Explanation:

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

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

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

$r_{1} = \sqrt{8^{2} \pm 9^{2}}) = \sqrt{145}$ $r_1=sqrt(8^2+-9^2))=sqrt 145$
$r_{2} = \sqrt{4^{2} + 6^{2}} = \sqrt{52}$ $r_2=sqrt(4^2+ 6^2) =sqrt 52$

$θ_{1} = {tan}^{- 1} (- \frac{9}{8}) \approx {311.63}^{\circ}, IV quadrant$ $theta_1=tan^-1(-9/8)~~ 311.63^@, " IV quadrant"$
$θ_{2} = {tan}^{- 1} (\frac{6}{4}) \approx {56.31}^{\circ}, I quadrant$ $theta_2=tan^-1(6/4)~~ 56.31^@, " I quadrant"$

$z_{1} + z_{2} = \sqrt{145} cos (311.63) + \sqrt{52} cos (56.31) + i (\sqrt{145} sin (311.63) + \sqrt{52} sin (56.31))$ $z_1 + z_2 = sqrt 145 cos(311.63) + sqrt 52 cos(56.31) + i (sqrt 145 sin(311.63)+ sqrt 52 sin(56.31))$

$\Rightarrow 8 + 4 + i (- 9 + 6)$ $=> 8 + 4 + i (-9 + 6 )$
$\Rightarrow 12 - 3 i$ $color(gray)(=> 12 - 3 i$

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How do you add $(8 - 9 i) + (4 + 6 i)$ $(8-9i)+(4+6i)$ in trigonometric form?

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How do you add $(8 - 9 i) + (4 + 6 i)$ $(8-9i)+(4+6i)$ in trigonometric form?