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 · 2018-06-25T09:59:49

$\Rightarrow 11 + i$ $color(crimson)(=> 11+ i$

Explanation:

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

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

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

$r_{1} = \sqrt{4^{2} + 8^{2}}) = \sqrt{80}$ $r_1=sqrt(4^2+ 8^2))=sqrt 80$
$r_{2} = \sqrt{3^{2} + - 3^{2}} = \sqrt{18}$ $r_2=sqrt(3^2+ -3^2) =sqrt 18$

$θ_{1} = {tan}^{- 1} (\frac{4}{8}) \approx {26.57}^{\circ}, I quadrant$ $theta_1=tan^-1(4 / 8)~~ 26.57^@, " I quadrant"$
$θ_{2} = {tan}^{- 1} (- \frac{3}{3}) \approx 315^{\circ}, IV quadrant$ $theta_2=tan^-1(-3/ 3)~~ 315^@, " IV quadrant"$

$z_{1} + z_{2} = \sqrt{80} cos (26.57) + \sqrt{18} cos (315) + i (\sqrt{80} sin 26.57 + \sqrt{18} sin 315)$ $z_1 + z_2 = sqrt 80 cos(26.57) + sqrt 18 cos(315) + i (sqrt 80 sin 26.57 + sqrt 18 sin 315)$

$\Rightarrow 8 + 3 + i (4 - 3)$ $=> 8 + 3 + i (4 - 3 )$

$\Rightarrow 11 + i$ $color(crimson)(=> 11+ i$

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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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How do you add (8+4i)(8+4i) and (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?