How do you divide $\frac{i + 2}{9 i + 14}$ $(i+2) / (9i+14)$ in trigonometric form?

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How do you divide $\frac{i + 2}{9 i + 14}$ $(i+2) / (9i+14)$ in trigonometric form?

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Answer 1 · 2018-04-19T07:57:18

$0.134 - 0.015 i$ $0.134-0.015i$

Explanation:

For a complex number $z = a + b i$ $z=a+bi$ it can be represented as $z = r (cos θ + i sin θ)$ $z=r(costheta+isintheta)$ where $r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$ and $θ = {tan}^{- 1} (\frac{b}{a})$ $theta=tan^-1(b/a)$

$\frac{2 + i}{14 + 9 i} = \frac{\sqrt{2^{2} + 1^{2}} (cos ({tan}^{- 1} (\frac{1}{2})) + i sin ({tan}^{- 1} (\frac{1}{2})))}{\sqrt{14^{2} + 9^{2}} (cos ({tan}^{- 1} (\frac{9}{14})) + i sin ({tan}^{- 1} (\frac{9}{14})))} \approx \frac{\sqrt{5} (cos (0.46) + i sin (0.46))}{\sqrt{277} (cos (0.57) + i sin (0.57))}$ $(2+i)/(14+9i)=(sqrt(2^2+1^2)(cos(tan^-1(1/2))+isin(tan^-1(1/2))))/(sqrt(14^2+9^2)(cos(tan^-1(9/14))+isin(tan^-1(9/14))))~~(sqrt5(cos(0.46)+isin(0.46)))/(sqrt277(cos(0.57)+isin(0.57)))$

Given $z_{1} = r_{1} ({cos θ}_{1} + i {sin θ}_{1})$ $z_1=r_1(costheta_1+isintheta_1)$ and $z_{2} = r_{2} ({cos θ}_{2} + i {sin θ}_{2})$ $z_2=r_2(costheta_2+isintheta_2)$ , $\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} (cos (θ_{1} - θ_{2}) + i sin (θ_{1} - θ_{2}))$ $z_1/z_2=r_1/r_2(cos(theta_1-theta_2)+isin(theta_1-theta_2))$

$\frac{z_{1}}{z_{2}} = \frac{\sqrt{5}}{\sqrt{277}} (cos (0.46 - 0.57) + i sin (0.46 - 0.57)) = \frac{\sqrt{1385}}{277} (cos (- 0.11) + i sin (- 0.11)) \approx \frac{\sqrt{1385}}{277} (0.99 - 0.11 i) \approx 0.134 - 0.015 i$ $z_1/z_2=sqrt5/sqrt277(cos(0.46-0.57)+isin(0.46-0.57))=sqrt1385/277(cos(-0.11)+isin(-0.11))~~sqrt1385/277(0.99-0.11i)~~0.134-0.015i$

Proof:
$\frac{2 + i}{14 + 9 i} \cdot \frac{14 - 9 i}{14 - 9 i} = \frac{28 - 4 i + 9}{14^{2} + 9^{2}} = \frac{37 - 4 i}{277} \approx 0.134 - 0.014 i$ $(2+i)/(14+9i)*(14-9i)/(14-9i)=(28-4i+9)/(14^2+9^2)=(37-4i)/277~~0.134-0.014i$

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How do you divide $\frac{i + 2}{9 i + 14}$ $(i+2) / (9i+14)$ in trigonometric form?

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