How do you add $(7 + 5 i) + (2 - 4 i)$ $(7+5i)+(2-4i)$ in trigonometric form?

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How do you add $(7 + 5 i) + (2 - 4 i)$ $(7+5i)+(2-4i)$ in trigonometric form?

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Answer 1 · 2018-06-13T13:18:36

$\sqrt{74} cos (0.6202) + \sqrt{20} cos (5.1760) + i (\sqrt{74} sin (0.6202) + \sqrt{20} sin (5.1760))$ $sqrt74cos(0.6202)+sqrt20cos(5.1760)+i(sqrt74sin(0.6202)+sqrt20sin(5.1760))$
$9 + i$ $9+i$

Explanation:

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

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

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

$r_{1} = \sqrt{7^{2} + 5^{2}} = \sqrt{49 + 25} = \sqrt{74}$ $r_1=sqrt(7^2+5^2)=sqrt(49+25)=sqrt74$
$r_{2} = \sqrt{2^{2} + 4^{2}} = \sqrt{4 + 16} = \sqrt{20}$ $r_2=sqrt(2^2+4^2)=sqrt(4+16)=sqrt20$

$θ_{1} = {tan}^{- 1} (\frac{5}{7}) \approx {0.6202}^{c}$ $theta_1=tan^-1(5/7)~~0.6202^c$
$θ_{2} = {tan}^{- 1} (- 2) \approx - {1.1071}^{c}$ $theta_2=tan^-1(-2)~~-1.1071^c$

However, as $2 - 4 i$ $2-4i$ is in quadrant 4, we need to add $2 π$ $2pi$ to get a positive angle variant.

$θ_{2} = 2 π + {tan}^{- 1} (- 2) \approx {5.1760}^{c}$ $theta_2=2pi+tan^-1(-2)~~5.1760^c$

$\sqrt{74} cos (0.6202) + \sqrt{20} cos (5.1760) + i (\sqrt{74} sin (0.6202) + \sqrt{20} sin (5.1760))$ $sqrt74cos(0.6202)+sqrt20cos(5.1760)+i(sqrt74sin(0.6202)+sqrt20sin(5.1760))$
$7 + 2 + i (5 - 4)$ $7+2+i(5-4)$
$9 + i$ $9+i$

Proof:

$7 + 5 i + 2 - 4$ $7+5i+2-4$
$(7 + 2) + i (5 - 4)$ $(7+2)+i(5-4)$
$9 + i$ $9+i$

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How do you add $(7 + 5 i) + (2 - 4 i)$ $(7+5i)+(2-4i)$ in trigonometric form?

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