How do you add $(8 - 2 i) + (7 - 3 i)$ $(8-2i)+(7-3i)$ in trigonometric form?

Question

How do you add $(8 - 2 i) + (7 - 3 i)$ $(8-2i)+(7-3i)$ in trigonometric form?

1 Answer

Impact of this question

1513 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-25T08:41:12

$\Rightarrow 15 - 5 i$ $color(cyan)(=> 15 - 5 i$

Explanation:

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

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

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

$r_{1} = \sqrt{8^{2} \pm 2^{2}}) = \sqrt{68}$ $r_1=sqrt(8^2+-2^2))=sqrt 68$
$r_{2} = \sqrt{7^{2} \pm 3^{2}} = \sqrt{58}$ $r_2=sqrt(7^2+-3^2) =sqrt 58$

$θ_{1} = {tan}^{- 1} (- \frac{2}{8}) \approx {345.96}^{\circ}, IV quadrant$ $theta_1=tan^-1(-2/8)~~ 345.96^@, " IV quadrant"$
$θ_{2} = {tan}^{- 1} (- \frac{3}{7}) \approx {336.8}^{\circ}, IV quadrant$ $theta_2=tan^-1(-3/ 7)~~ 336.8^@, " IV quadrant"$

$z_{1} + z_{2} = \sqrt{68} cos (345.96) + \sqrt{58} cos (336.8) + i (\sqrt{68} sin 345.96 + \sqrt{58} sin 336.8)$ $z_1 + z_2 = sqrt 68 cos(345.96) + sqrt 58 cos(336.8) + i (sqrt 68 sin 345.96 + sqrt 58 sin 336.8)$

$\Rightarrow 8 + 7 + i (- 2 - 3)$ $=> 8 + 7 + i (-2 - 3 )$

$\Rightarrow 15 - 5 i$ $color(cyan)(=> 15 - 5 i$

Science

Math

Humanities

... and beyond

How do you add $(8 - 2 i) + (7 - 3 i)$ $(8-2i)+(7-3i)$ in trigonometric form?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you add (8−2i)+(7−3i)(8-2i)+(7-3i) in trigonometric form?

1 Answer

Explanation:

Related questions

Impact of this question

How do you add $(8 - 2 i) + (7 - 3 i)$ $(8-2i)+(7-3i)$ in trigonometric form?