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

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

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Answer 1 · 2016-11-03T05:54:16

$\frac{2 i - 7}{3 i - 2} = \sqrt{\frac{53}{13}} (cos θ + i sin θ)$ $(2i-7)/(3i-2)=sqrt(53/13)(costheta+isintheta)$ , where $θ = {tan}^{- 1} (\frac{17}{20})$ $theta=tan^(-1)(17/20)$

Explanation:

Let us write the two complex numbers in polar coordinates and let them be

$z_{1} = r_{1} (cos α + i sin α)$ $z_1=r_1(cosalpha+isinalpha)$ and $z_{2} = r_{2} (cos β + i sin β)$ $z_2=r_2(cosbeta+isinbeta)$

Here, if two complex numbers are $a_{1} + i b_{1}$ $a_1+ib_1$ and $a_{2} + i b_{2}$ $a_2+ib_2$ $r_{1} = \sqrt{a_{1}^{2} + b_{1}^{2}}$ $r_1=sqrt(a_1^2+b_1^2)$ , $r_{2} = \sqrt{a_{2}^{2} + b_{2}^{2}}$ $r_2=sqrt(a_2^2+b_2^2)$ and $α = {tan}^{- 1} (\frac{b_{1}}{a_{1}})$ $alpha=tan^(-1)(b_1/a_1)$ , $β = {tan}^{- 1} (\frac{b_{2}}{a_{2}})$ $beta=tan^(-1)(b_2/a_2)$

Their division leads us to

${\frac{r_{1}}{r_{2}}} {\frac{cos α + i sin α}{cos β + i sin β}}$ ${r_1/r_2}{(cosalpha+isinalpha)/(cosbeta+isinbeta)}$ or

${\frac{r_{1}}{r_{2}}} {\frac{cos α + i sin α}{cos β + i sin β} \times \frac{cos β - i sin β}{cos β - i sin β}}$ ${r_1/r_2}{(cosalpha+isinalpha)/(cosbeta+isinbeta)xx(cosbeta-isinbeta)/(cosbeta-isinbeta)}$

$(\frac{r_{1}}{r_{2}}) \frac{(cos α cos β + sin α sin β) + i (sin α cos β - cos α sin β)}{({cos}^{2} β + {sin}^{2} β)}$ $(r_1/r_2){(cosalphacosbeta+sinalphasinbeta)+i(sinalphacosbeta-cosalphasinbeta))/((cos^2beta+sin^2beta))$ or

$(\frac{r_{1}}{r_{2}}) \cdot (cos (α - β) + i sin (α - β))$ $(r_1/r_2)*(cos(alpha-beta)+isin(alpha-beta))$ or

$\frac{z_{1}}{z_{2}}$ $z_1/z_2$ is given by $(\frac{r_{1}}{r_{2}}, (α - β))$ $(r_1/r_2, (alpha-beta))$

So for division complex number $z_{1}$ $z_1$ by $z_{2}$ $z_2$ , take new angle as $(α - β)$ $(alpha-beta)$ and modulus the ratio $\frac{r_{1}}{r_{2}}$ $r_1/r_2$ of the modulus of two numbers.

Here $2 i - 7 = - 7 + 2 i$ $2i-7=-7+2i$ can be written as $r_{1} (cos α + i sin α)$ $r_1(cosalpha+isinalpha)$ where $r_{1} = \sqrt{{(- 7)}^{2} + 2^{2}} = \sqrt{53}$ $r_1=sqrt((-7)^2+2^2)=sqrt53$ and $α = {tan}^{- 1} (- \frac{2}{7})$ $alpha=tan^(-1)(-2/7)$

and $3 i - 2 = - 2 + 3 i$ $3i-2=-2+3i$ can be written as $r_{2} (cos β + i sin β)$ $r_2(cosbeta+isinbeta)$ where $r_{2} = \sqrt{{(- 2)}^{2} + 3^{2}} = \sqrt{13}$ $r_2=sqrt((-2)^2+3^2)=sqrt13$ and $β = {tan}^{- 1} (- \frac{3}{2})$ $beta=tan^(-1)(-3/2)$

and $\frac{z_{1}}{z_{2}} = \frac{\sqrt{53}}{\sqrt{13}} (cos θ + i sin θ)$ $z_1/z_2=sqrt53/sqrt13(costheta+isintheta)$ , where $θ = α - β$ $theta=alpha-beta$

Hence, $tan θ = tan (α - β) = \frac{tan α - tan β}{1 + tan α tan β} = \frac{(- \frac{2}{7}) - (- \frac{3}{2})}{1 + (- \frac{2}{7}) \times (- \frac{3}{2})} = \frac{- \frac{2}{7} + \frac{3}{2}}{1 + \frac{3}{7}} = \frac{\frac{17}{14}}{\frac{10}{7}} = \frac{17}{14} \times \frac{7}{10} = \frac{17}{20}$ $tantheta=tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)=((-2/7)-(-3/2))/(1+(-2/7)xx(-3/2))=(-2/7+3/2)/(1+3/7)=(17/14)/(10/7)=17/14xx7/10=17/20$ .

Hence, $\frac{2 i - 7}{3 i - 2} = \sqrt{\frac{53}{13}} (cos θ + i sin θ)$ $(2i-7)/(3i-2)=sqrt(53/13)(costheta+isintheta)$ , where $θ = {tan}^{- 1} (\frac{17}{20})$ $theta=tan^(-1)(17/20)$

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