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

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

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Answer 1 · 2018-06-25T08:29:57

$\Rightarrow - 2 - 10 i$ $color(cyan)(=> -2 - 10 i$

Explanation:

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

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

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

$r_{1} = \sqrt{- 1^{2} \pm 8^{2}}) = \sqrt{65}$ $r_1=sqrt(-1^2+-8^2))=sqrt 65$
$r_{2} = \sqrt{- 1^{2} \pm 2^{2}} = \sqrt{5}$ $r_2=sqrt(-1^2+-2^2) =sqrt 5$

$θ_{1} = {tan}^{- 1} (- \frac{8}{- 1}) \approx {262.87}^{\circ}, III quadrant$ $theta_1=tan^-1(-8/-1)~~ 262.87^@, " III quadrant"$
$θ_{2} = {tan}^{- 1} (- \frac{2}{- 1}) \approx {243.43}^{\circ}, III quadrant$ $theta_2=tan^-1(-2/-1)~~ 243.43^@, " III quadrant"$

$z_{1} + z_{2} = \sqrt{65} cos (262.87) + \sqrt{5} cos (243.43) + i (\sqrt{65} sin (262.87) + \sqrt{5} sin (243.43))$ $z_1 + z_2 = sqrt 65 cos(262.87) + sqrt 5 cos(243.43) + i (sqrt 65 sin(262.87)+ sqrt 5 sin(243.43))$

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

$\Rightarrow - 2 - 10 i$ $color(cyan)(=> -2 - 10 i$

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

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

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Explanation:

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