How do you add $(- 9 + 2 i) + (2 + 4 i)$ $(-9+2i)+(2+4i)$ in trigonometric form?

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How do you add $(- 9 + 2 i) + (2 + 4 i)$ $(-9+2i)+(2+4i)$ in trigonometric form?

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Answer 1 · 2018-07-27T05:44:21

$\Rightarrow - 7 + 6 i, II Quadrant$ $color(magenta)(=> -7 + 6 i, " II Quadrant"$

Explanation:

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

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

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

$r_{1} = \sqrt{- 9^{2} + 2^{2}}) = \sqrt{85}$ $r_1=sqrt(-9^2+ 2^2))=sqrt 85$
$r_{2} = \sqrt{2^{2} + 4^{2}} = \sqrt{20}$ $r_2=sqrt(2^2+ 4^2) =sqrt 20$

$θ_{1} = {tan}^{- 1} (\frac{2}{- 9}) \approx {167.4712}^{\circ}, II quadrant$ $theta_1=tan^-1(2 / -9) ~~ 167.4712^@, " II quadrant"$
$θ_{2} = {tan}^{- 1} (\frac{4}{2}) \approx {63.4349}^{\circ}, I quadrant$ $theta_2=tan^-1(4/ 2)~~ 63.4349^@, " I quadrant"$

$z_{1} + z_{2} = \sqrt{85} cos (167.4712) + \sqrt{20} cos (63.4349) + i (\sqrt{85} sin 167.4712 + \sqrt{20} sin 63.4349)$ $z_1 + z_2 = sqrt 85 cos(167.4712) + sqrt 20 cos(63.4349) + i (sqrt 85 sin 167.4712 + sqrt 20 sin 63.4349)$

$\Rightarrow - 9 + 2 + i (2 + 4)$ $=> -9 + 2 + i (2 + 4 )$

$\Rightarrow - 7 + 6 i, II Quadrant$ $color(magenta)(=> -7 + 6 i, " II Quadrant"$

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How do you add $(- 9 + 2 i) + (2 + 4 i)$ $(-9+2i)+(2+4i)$ in trigonometric form?

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How do you add (−9+2i)+(2+4i)(-9+2i)+(2+4i) in trigonometric form?

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How do you add $(- 9 + 2 i) + (2 + 4 i)$ $(-9+2i)+(2+4i)$ in trigonometric form?