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

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

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Answer 1 · 2016-04-28T08:38:11

$\frac{- i + 2}{2 i + 4} = \frac{1}{2} (cos θ + i sin θ)$ $(-i+2)/(2i+4)=1/2(costheta+isintheta)$ , where $θ = {tan}^{- 1} (- \frac{4}{3})$ $theta=tan^(-1)(-4/3)$

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 $- i + 2$ $-i+2$ can be written as $r_{1} (cos α + i sin α)$ $r_1(cosalpha+isinalpha)$ where $r_{1} = \sqrt{2^{2} + {(- 1)}^{2}} = \sqrt{5}$ $r_1=sqrt(2^2+(-1)^2)=sqrt5$ and $α = {tan}^{- 1} (- \frac{1}{2})$ $alpha=tan^(-1)(-1/2)$

and $2 i + 4$ $2i+4$ can be written as $r_{2} (cos β + i sin β)$ $r_2(cosbeta+isinbeta)$ where $r_{2} = \sqrt{4^{2} + 2^{2}} = \sqrt{20} = 2 \sqrt{5}$ $r_2=sqrt(4^2+2^2)=sqrt20=2sqrt5$ and $β = {tan}^{- 1} (\frac{2}{4}) = {tan}^{- 1} (\frac{1}{2})$ $beta=tan^(-1)(2/4)=tan^(-1)(1/2)$

and $\frac{z_{1}}{z_{2}} = \frac{\sqrt{5}}{2 \sqrt{5}} (cos θ + i sin θ)$ $z_1/z_2=sqrt5/(2sqrt5)(costheta+isintheta)$ , where $θ = α - β$ $theta=alpha-beta$

Hence, $tan θ = tan (α - β) = \frac{tan α - tan β}{1 + tan α tan β} = \frac{(- \frac{1}{2}) - (\frac{1}{2})}{1 + (- \frac{1}{2}) \times (\frac{1}{2})} = \frac{- 1}{\frac{3}{4}} = - \frac{4}{3}$ $tantheta=tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)=((-1/2)-(1/2))/(1+(-1/2)xx(1/2))=(-1)/(3/4)=-4/3$ .

Hence, $\frac{- i + 2}{2 i + 4} = \frac{1}{2} (cos θ + i sin θ)$ $(-i+2)/(2i+4)=1/2(costheta+isintheta)$ , where $θ = {tan}^{- 1} (- \frac{4}{3})$ $theta=tan^(-1)(-4/3)$

Answer 2 · 2016-06-30T05:09:27

$\frac{- i + 2}{2 i + 4} = \frac{1}{2} (cos θ + sin θ)$ $(-i+2)/(2i+4)=1/2(costheta+sintheta)$ ,where $θ = {tan}^{- 1} (- \frac{4}{3})$ $theta=tan^-1(-4/3)$

Explanation:

$\frac{- i + 2}{2 i + 4}$ $(-i+2)/(2i+4)$

$= \frac{1}{2} \cdot \frac{(2 - i) (2 - i)}{(2 + i) (2 - i)}$ $=1/2*((2-i)(2-i))/((2+i)(2-i))$

$= \frac{1}{2} \cdot \frac{2^{2} + i^{2} - 2 \cdot 2 i}{2^{2} - i^{2}}$ $=1/2*(2^2+i^2-2*2i)/(2^2-i^2)$

$= \frac{1}{2} \cdot \frac{4 - 1 - 4 i}{4 - (- 1)}$ $=1/2*(4-1-4i)/(4-(-1))$

$= \frac{1}{10} (3 - 4 i)$ $=1/10(3-4i)$

Now $\sqrt{3^{2} + 4^{2}} = 5$ $sqrt(3^2+4^2)=5$ ,So

The given expression

$= \frac{1}{2} (\frac{3}{5} - \frac{4}{5} i)$ $=1/2(3/5-4/5i)$

Now if we take $\frac{3}{5} = cos θ$ $3/5=costheta$ and $- \frac{4}{5} = sin θ$ $-4/5=sintheta$
i.e. $tan θ = - \frac{4}{3}$ $tantheta=-4/3$
then we can write

$\frac{- i + 2}{2 i + 4} = \frac{1}{2} (cos θ + sin θ)$ $(-i+2)/(2i+4)=1/2(costheta+sintheta)$ ,where $θ = {tan}^{- 1} (- \frac{4}{3})$ $theta=tan^-1(-4/3)$

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