How do you divide $\frac{i + 2}{7 i + 3}$ $( i+2) / (7i +3)$ in trigonometric form?

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How do you divide $\frac{i + 2}{7 i + 3}$ $( i+2) / (7i +3)$ in trigonometric form?

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Answer 1 · 2018-01-05T11:53:38

$\frac{\sqrt{290}}{58} (cos (40.24) - i sin (40.24))$ $sqrt(290)/58(cos(40.24)-isin(40.24))$

Explanation:

For a complex number $z = a + b i$ $z=a+bi$ , we can rewrite it in the form $z = r (cos θ + i sin θ)$ $z=r(costheta+isintheta)$ , where $r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$ and $θ = {tan}^{- 1} (\frac{b}{a})$ $theta=tan^(-1)(b/a)$

$z_{1} = 2 + i$ $z_1=2+i$
$r_{1} = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$ $r_1=sqrt(2^2+1^2)=sqrt(5)$
$θ_{1} = {tan}^{- 1} (\frac{1}{2})$ $theta_1=tan^(-1)(1/2)$
$z_{1} = \sqrt{5} (cos ({tan}^{- 1} (\frac{1}{2})) + i sin ({tan}^{- 1} (\frac{1}{2}))$ $z_1=sqrt(5)(cos(tan^(-1)(1/2))+isin(tan^(-1)(1/2))$

$z_{2} = 3 + 7 i$ $z_2=3+7i$
$r_{2} = \sqrt{3^{2} + 7^{2}} = \sqrt{58}$ $r_2=sqrt(3^2+7^2)=sqrt(58)$
$θ_{2} = {tan}^{- 1} (\frac{7}{3})$ $theta_2=tan^(-1)(7/3)$
$z_{2} = \sqrt{5} (cos ({tan}^{- 1} (\frac{7}{3})) + i sin ({tan}^{- 1} (\frac{7}{3}))$ $z_2=sqrt(5)(cos(tan^(-1)(7/3))+isin(tan^(-1)(7/3))$

$\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} (cos (θ_{1} - θ_{2}) + i sin (θ_{1} - θ_{2}))$ $z_1/z_2=r_1/r_2(cos(theta_1-theta_2)+isin(theta_1-theta_2))$

$\frac{z_{1}}{z_{2}} = \frac{\sqrt{5}}{\sqrt{58}} (cos ({tan}^{- 1} (\frac{1}{2}) - {tan}^{- 1} (\frac{7}{3})) + i sin ({tan}^{- 1} (\frac{1}{2}) - {tan}^{- 1} (\frac{7}{3})))$ $z_1/z_2=sqrt(5)/sqrt(58)(cos(tan^(-1)(1/2)-tan^(-1)(7/3))+isin(tan^(-1)(1/2)-tan^(-1)(7/3)))$
$\approx \frac{\sqrt{290}}{58} (cos (- 40.24) + i sin (- 40.24))$ $~~sqrt(290)/58(cos(-40.24)+isin(-40.24))$

Since $cos (x) = cos (- x)$ $cos(x)=cos(-x)$ and $sin (- x) = - sin (x)$ $sin(-x)=-sin(x)$

$= \frac{\sqrt{290}}{58} (cos (40.24) - i sin (40.24))$ $=sqrt(290)/58(cos(40.24)-isin(40.24))$

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