How do you divide $\frac{8 - 2 i}{5 - 3 i}$ $(8-2i) / (5-3i)$ in trigonometric form?

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How do you divide $\frac{8 - 2 i}{5 - 3 i}$ $(8-2i) / (5-3i)$ in trigonometric form?

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Answer 1 · 2016-01-27T09:26:39

$\sqrt{2} (cos ({tan}^{- 1} (\frac{7}{23})) + i sin ({tan}^{- 1} (\frac{7}{23}))$ $sqrt2(cos(tan^-1 (7/23))+isin(tan^-1 (7/23))$ OR

$\sqrt{2} (cos ({16.9275}^{\circ}) + i sin ({16.9275}^{\circ}))$ $sqrt2(cos(16.9275^@)+isin(16.9275^@))$

Explanation:

Start from the given complex number

$\frac{8 - 2 i}{5 - 3 i}$ $(8-2i)/(5-3i)$

Convert the numerator

$r_{1} = \sqrt{8^{2} + {(- 2)}^{2}} = \sqrt{68}$ $r_1=sqrt(8^2+(-2)^2)=sqrt68$

$θ_{1} = {tan}^{- 1} (\frac{- 2}{8}) = {tan}^{- 1} (\frac{- 1}{4})$ $theta_1=tan^-1 ((-2)/8)=tan^-1 ((-1)/4)$

then

$8 - 2 i = \sqrt{68} [cos ({tan}^{- 1} (\frac{- 1}{4})) + i sin ({tan}^{- 1} (\frac{- 1}{4}))]$ $8-2i=sqrt68[cos(tan^-1 ((-1)/4))+isin(tan^-1 ((-1)/4))]$

Convert the denominator

$r_{2} = \sqrt{5^{2} + {(- 3)}^{2}} = \sqrt{34}$ $r_2=sqrt(5^2+(-3)^2)=sqrt34$

$θ_{2} = {tan}^{- 1} (\frac{- 3}{5}) = {tan}^{- 1} (\frac{- 3}{5})$ $theta_2=tan^-1 ((-3)/5)=tan^-1 ((-3)/5)$

then

$5 - 3 i = \sqrt{34} [cos ({tan}^{- 1} (\frac{- 3}{5})) + i sin ({tan}^{- 1} (\frac{- 3}{5}))]$ $5-3i=sqrt34[cos(tan^-1 ((-3)/5))+isin(tan^-1 ((-3)/5))]$

Let us divide now, from the given with the equivalent

$\frac{8 - 2 i}{5 - 3 i} = \frac{\sqrt{68} [cos ({tan}^{- 1} (\frac{- 1}{4})) + i sin ({tan}^{- 1} (\frac{- 1}{4}))]}{\sqrt{34} [cos ({tan}^{- 1} (\frac{- 3}{5})) + i sin ({tan}^{- 1} (\frac{- 3}{5}))]}$ $(8-2i)/(5-3i)=(sqrt68[cos(tan^-1 ((-1)/4))+isin(tan^-1 ((-1)/4))])/(sqrt34[cos(tan^-1 ((-3)/5))+isin(tan^-1 ((-3)/5))])$

Divide using the following formula

$(\frac{r_{1}}{r_{2}}) \cdot [cos (θ_{1} - θ_{2}) + i sin (θ 1 - θ_{2})]$ $(r_1/r_2)*[cos (theta_1-theta_2)+isin(theta1-theta_2)]$

$\frac{8 - 2 i}{5 - 3 i} = \sqrt{\frac{68}{34}} [cos ({tan}^{- 1} (\frac{- 1}{4}) - {tan}^{- 1} (\frac{- 3}{5})) + i sin ({tan}^{- 1} (\frac{- 1}{4}) - {tan}^{- 1} (\frac{- 3}{5}))]$ $(8-2i)/(5-3i)= sqrt(68/34)[cos(tan^-1 ((-1)/4)-tan^-1 ((-3)/5))+isin(tan^-1 ((-1)/4)-tan^-1 ((-3)/5))]$

Take note: that
$tan (θ_{1} - θ_{2}) = \frac{{tan θ}_{1} - {tan θ}_{2}}{1 + {tan θ}_{1} \cdot {tan θ}_{2}} = \frac{- \frac{1}{4} - (- \frac{3}{5})}{1 + (- \frac{1}{4}) \cdot (- \frac{3}{5})}$ $tan (theta_1-theta_2)=(tan theta_1-tan theta_2) /(1+tan theta_1*tan theta_2)=(-1/4-(-3/5))/(1+(-1/4)*(-3/5))$

$tan (θ_{1} - θ_{2}) = \frac{7}{23}$ $tan (theta_1-theta_2)=7/23$

$θ_{1} - θ_{2} = {tan}^{- 1} (\frac{7}{23})$ $theta_1-theta_2=tan^-1 (7/23)$

and

$\frac{8 - 2 i}{5 - 3 i} = \sqrt{2} (cos ({tan}^{- 1} (\frac{7}{23})) + i sin ({tan}^{- 1} (\frac{7}{23}))$ $(8-2i)/(5-3i)=sqrt2(cos(tan^-1 (7/23))+isin(tan^-1 (7/23))$

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How do you divide $\frac{8 - 2 i}{5 - 3 i}$ $(8-2i) / (5-3i)$ in trigonometric form?

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