How do you add $(- 2 + 3 i) + (6 + i)$ $(-2+3i)+(6+i)$ in trigonometric form?

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How do you add $(- 2 + 3 i) + (6 + i)$ $(-2+3i)+(6+i)$ in trigonometric form?

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Answer 1 · 2018-06-25T07:00:54

$(- 2 + 3 i) + (6 + i) = 4 + 4 i = 4 (1 + i)$ $color(crimson)((-2 + 3 i) + (6 + i) = 4 + 4 i = 4(1 + i)$

Explanation:

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

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

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

$r_{1} = \sqrt{- 2^{2} + 3^{2}}) = \sqrt{13}$ $r_1=sqrt(-2^2+3^2))=sqrt 13$
$r_{2} = \sqrt{6^{2} \pm 1^{2}} = \sqrt{37}$ $r_2=sqrt(6^2+-1^2) =sqrt 37$

$θ_{1} = {tan}^{- 1} (\frac{3}{- 2}) \approx {123.69}^{\circ}, II quadrant$ $theta_1=tan^-1(3/-2)~~ 123.69^@, " II quadrant"$
$θ_{2} = {tan}^{- 1} (\frac{1}{6}) \approx {9.4623}^{\circ}, I quadrant$ $theta_2=tan^-1(1/6)~~ 9.4623^@, " I quadrant"$

$z_{1} + z_{2} = \sqrt{13} cos (123.69) + \sqrt{37} cos (9.4623) + i (\sqrt{13} sin (123.69) + \sqrt{37} sin (9.4623))$ $z_1 + z_2 = sqrt13 cos(123.69) + sqrt37 cos(9.4623) + i (sqrt 13 sin(123.69)+ sqrt 37 sin(9.4623))$

$\Rightarrow - 2 + 6 + i (3 + 1)$ $=> -2 + 6 + i (3 + 1 )$
$\Rightarrow 4 + 4 i$ $color(brown)(=> 4 + 4 i$

Proof:

$- 2 + 3 i + 6 + i$ $-2 + 3i + 6 + i$

$\Rightarrow 4 + 4 i$ $=> 4 + 4 i$

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How do you add $(- 2 + 3 i) + (6 + i)$ $(-2+3i)+(6+i)$ in trigonometric form?

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Explanation:

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How do you add (−2+3i)+(6+i)(-2+3i)+(6+i) in trigonometric form?

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How do you add $(- 2 + 3 i) + (6 + i)$ $(-2+3i)+(6+i)$ in trigonometric form?