How do you add $(2 - 5 i)$ $(2-5i)$ and $(- 2 - 2 i)$ $(-2-2i)$ in trigonometric form?

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How do you add $(2 - 5 i)$ $(2-5i)$ and $(- 2 - 2 i)$ $(-2-2i)$ in trigonometric form?

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Answer 1 · 2018-07-09T11:40:21

$\Rightarrow - 7 i$ $color(red)(=> - 7i$

Explanation:

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

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

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

$r_{1} = \sqrt{2^{2} + - 5^{2}}) = \sqrt{29}$ $r_1=sqrt(2^2+ -5^2))=sqrt 29$
$r_{2} = \sqrt{- 2^{2} + - 2^{2}} = \sqrt{8}$ $r_2=sqrt(-2^2+ -2^2) =sqrt 8$

$θ_{1} = {tan}^{- 1} (- \frac{5}{2}) \approx {291.8}^{\circ}, IV quadrant$ $theta_1=tan^-1(-5 / 2)~~ 291.8^@, " IV quadrant"$
$θ_{2} = {tan}^{- 1} (- \frac{2}{- 2}) \approx 225^{\circ}, III quadrant$ $theta_2=tan^-1(-2/ -2)~~ 225^@, " III quadrant"$

$z_{1} + z_{2} = \sqrt{29} cos (291.8) + \sqrt{8} cos (225) + i (\sqrt{29} sin 291.8 + \sqrt{8} sin 225)$ $z_1 + z_2 = sqrt 29 cos(291.8) + sqrt 8 cos(225) + i (sqrt 29 sin 291.8 + sqrt 8 sin 225)$

$\Rightarrow 2 - 2 + i (- 5 - 2)$ $=> 2 - 2 + i (-5 - 2 )$

$\Rightarrow - 7 i$ $color(red)(=> - 7i$

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How do you add $(2 - 5 i)$ $(2-5i)$ and $(- 2 - 2 i)$ $(-2-2i)$ in trigonometric form?

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Explanation:

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How do you add $(2 - 5 i)$ $(2-5i)$ and $(- 2 - 2 i)$ $(-2-2i)$ in trigonometric form?