How do you divide $\frac{2 - 4 i}{5 + 8 i}$ $(2-4i)/(5+8i)$ in trigonometric form?

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How do you divide $\frac{2 - 4 i}{5 + 8 i}$ $(2-4i)/(5+8i)$ in trigonometric form?

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Answer 1 · 2016-07-04T07:58:57

$\frac{2 - 4 i}{5 + 8 i} = \sqrt{\frac{20}{89}} (cos θ + i sin θ)$ $(2-4i)/(5+8i)=sqrt(20/89)(costheta+isintheta)$ , where $θ = {tan}^{- 1} (\frac{18}{11})$ $theta=tan^(-1)(18/11)$

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 $2 - 4 i$ $2-4i$ can be written as $r_{1} (cos α + i sin α)$ $r_1(cosalpha+isinalpha)$ where $r_{1} = \sqrt{2^{2} + {(- 4)}^{2}} = \sqrt{20}$ $r_1=sqrt(2^2+(-4)^2)=sqrt20$ and $α = {tan}^{- 1} (- \frac{4}{2}) = {tan}^{- 1} (- 2)$ $alpha=tan^(-1)(-4/2)=tan^(-1)(-2)$

and $5 + 8 i$ $5+8i$ can be written as $r_{2} (cos β + i sin β)$ $r_2(cosbeta+isinbeta)$ where $r_{2} = \sqrt{5^{2} + 8^{2}} = \sqrt{89}$ $r_2=sqrt(5^2+8^2)=sqrt89$ and $β = {tan}^{- 1} (\frac{8}{5})$ $beta=tan^(-1)(8/5)$

and $\frac{z_{1}}{z_{2}} = \frac{\sqrt{20}}{\sqrt{89}} (cos θ + i sin θ)$ $z_1/z_2=sqrt20/(sqrt89)(costheta+isintheta)$ , where $θ = α - β$ $theta=alpha-beta$

Hence, $tan θ = tan (α - β) = \frac{tan α - tan β}{1 + tan α tan β} = \frac{(- 2) - (\frac{8}{5})}{1 + (- 2) \times (\frac{8}{5})} = \frac{- \frac{18}{5}}{- \frac{11}{5}} = \frac{18}{11}$ $tantheta=tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)=((-2)-(8/5))/(1+(-2)xx(8/5))=(-18/5)/(-11/5)=18/11$ .

Hence, $\frac{2 - 4 i}{5 + 8 i} = \sqrt{\frac{20}{89}} (cos θ + i sin θ)$ $(2-4i)/(5+8i)=sqrt(20/89)(costheta+isintheta)$ , where $θ = {tan}^{- 1} (\frac{18}{11})$ $theta=tan^(-1)(18/11)$

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How do you divide $\frac{2 - 4 i}{5 + 8 i}$ $(2-4i)/(5+8i)$ in trigonometric form?

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How do you divide $\frac{2 - 4 i}{5 + 8 i}$ $(2-4i)/(5+8i)$ in trigonometric form?