How do you add $(- 1 + 8 i) + (4 - 2 i)$ $(-1+8i)+(4-2i)$ in trigonometric form?

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How do you add $(- 1 + 8 i) + (4 - 2 i)$ $(-1+8i)+(4-2i)$ in trigonometric form?

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Answer 1 · 2018-06-25T10:58:11

$\Rightarrow 3 + 6 i$ $color(indigo)(=> 3+ 6 i$

Explanation:

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

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

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

$r_{1} = \sqrt{- 1^{2} + 8^{2}}) = \sqrt{65}$ $r_1=sqrt(-1^2+ 8^2))=sqrt 65$
$r_{2} = \sqrt{- 2^{2} + 4^{2}} = \sqrt{20}$ $r_2=sqrt(-2^2+ 4^2) =sqrt 20$

$θ_{1} = {tan}^{- 1} (\frac{8}{- 1}) \approx {97.13}^{\circ}, II quadrant$ $theta_1=tan^-1(8 / -1)~~ 97.13^@, " II quadrant"$
$θ_{2} = {tan}^{- 1} (- \frac{2}{4}) \approx {333.43}^{\circ}, IV quadrant$ $theta_2=tan^-1(-2/ 4)~~ 333.43^@, " IV quadrant"$

$z_{1} + z_{2} = \sqrt{65} cos (97.13) + \sqrt{20} cos (333.43) + i (\sqrt{65} sin 97.13 + \sqrt{20} sin 333.43)$ $z_1 + z_2 = sqrt 65 cos(97.13) + sqrt 20 cos(333.43) + i (sqrt 65 sin 97.13 + sqrt 20 sin 333.43)$

$\Rightarrow - 1 + 4 + i (8 - 2)$ $=> -1 + 4 + i (8 - 2 )$

$\Rightarrow 3 + 6 i$ $color(indigo)(=> 3+ 6 i$

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How do you add $(- 1 + 8 i) + (4 - 2 i)$ $(-1+8i)+(4-2i)$ in trigonometric form?

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How do you add (−1+8i)+(4−2i)(-1+8i)+(4-2i) in trigonometric form?

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How do you add $(- 1 + 8 i) + (4 - 2 i)$ $(-1+8i)+(4-2i)$ in trigonometric form?