How do you add $(1 + 7 i) + (- 4 - 2 i)$ $(1+7i)+(-4-2i)$ in trigonometric form?

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How do you add $(1 + 7 i) + (- 4 - 2 i)$ $(1+7i)+(-4-2i)$ in trigonometric form?

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Answer 1 · 2018-06-25T06:52:44

$\Rightarrow - 3 + 5 i$ $color(brown)(=> -3 + 5i$

Explanation:

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

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

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

$r_{1} = \sqrt{1^{2} + 7^{2}}) = \sqrt{50}$ $r_1=sqrt(1^2+7^2))=sqrt50$
$r_{2} = \sqrt{- 4^{2} \pm 2^{2}} = \sqrt{20}$ $r_2=sqrt(-4^2+-2^2) =sqrt20$

$θ_{1} = {tan}^{- 1} (\frac{7}{1}) \approx {81.87}^{\circ}, I quadrant$ $theta_1=tan^-1(7/1)~~81.87^@, " I quadrant"$
$θ_{2} = {tan}^{- 1} (- \frac{2}{- 4}) \approx {206.57}^{\circ}, III quadrant$ $theta_2=tan^-1(-2/-4)~~206.57^@, " III quadrant"$

$z_{1} + z_{2} = \sqrt{50} cos (81.87) + \sqrt{20} cos (206.57) + i (\sqrt{50} sin (81.87) + \sqrt{20} sin (206.57))$ $z_1 + z_2 = sqrt50 cos(81.87) + sqrt20 cos(206.57) + i (sqrt50 sin(81.87)+ sqrt20 sin(206.57))$

$\Rightarrow 1 - 4 + i (7 - 2)$ $=> 1 -4 + i (7- 2 )$
$\Rightarrow - 3 + 5 i$ $color(brown)(=> -3 + 5i$

Proof:

$1 + 7 i - 4 - 2 i$ $1 + 7i - 4 - 2i$

$\Rightarrow - 3 + 5 i$ $=> -3 + 5i$

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How do you add $(1 + 7 i) + (- 4 - 2 i)$ $(1+7i)+(-4-2i)$ in trigonometric form?

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