How do you divide $\frac{2 i - 6}{- 12 i + 4}$ $( 2i-6) / ( -12i +4 )$ in trigonometric form?

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How do you divide $\frac{2 i - 6}{- 12 i + 4}$ $( 2i-6) / ( -12i +4 )$ in trigonometric form?

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Answer 1 · 2016-05-19T07:43:14

$(2 i - 6) \div (- 12 i + 4) = \frac{1}{2} (cos ρ + i sin ρ)$ $(2i-6)-:(-12i+4)=1/2(cosrho+isinrho)$ where $ρ = {tan}^{- 1} (- \frac{5}{3})$ $rho=tan^(-1)(-5/3)$

Explanation:

Let us first write $(2 i - 6)$ $(2i-6)$ and $(- 12 i + 4)$ $(-12i+4)$ in trigonometric form.

$a + i b$ $a+ib$ can be written in trigonometric form $r e^{i θ} = r cos θ + i r sin θ = r (cos θ + i sin θ)$ $re^(itheta)=rcostheta+irsintheta=r(costheta+isintheta)$ ,
where $r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$ and $tan θ = \frac{b}{a}$ $tantheta=b/a$ or $θ = arctan (\frac{b}{a})$ $theta=arctan(b/a)$

Hence $2 i - 6 = (- 6 + 2 i) = \sqrt{{(- 6)}^{2} + 2^{2}} [cos α + i sin α]$ $2i-6=(-6+2i)=sqrt((-6)^2+2^2)[cosalpha+isinalpha]$ or

$\sqrt{40} e^{i α}$ $sqrt40e^(ialpha)$ , where $tan α = \frac{- 1}{3}$ $tanalpha=(-1)/3$ and

$- 12 i + 4 = (4 - 12 i) = \sqrt{4^{2} + {(- 12)}^{2}} [cos β + i sin β]$ $-12i+4=(4-12i)=sqrt(4^2+(-12)^2)[cosbeta+isinbeta]$ or

$\sqrt{160} e^{i β}$ $sqrt160e^(ibeta]$ , where $tan β = \frac{- 12}{4} = - 3$ $tanbeta=(-12)/4=-3$

Hence $(2 i - 6) \div (- 12 i + 4) = \frac{\sqrt{40} e^{i α}}{\sqrt{160} e^{i β}} = \sqrt{\frac{1}{4}} e^{i (α - β)} = \frac{e^{i (α - β)}}{2} = \frac{1}{2} (cos (α - β) + i sin (α - β))$ $(2i-6)-:(-12i+4)=(sqrt40e^(ialpha))/(sqrt160e^(ibeta])=sqrt(1/4)e^(i(alpha-beta))=e^(i(alpha-beta))/2=1/2(cos(alpha-beta)+isin(alpha-beta))$

Now, $tan (α - β) = \frac{tan α - tan β}{1 + tan α tan β}$ $tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)$

= $\frac{- \frac{1}{3} + (- 3)}{1 + (- \frac{1}{3}) \cdot (- 3)} = \frac{- \frac{10}{3}}{2} = - \frac{5}{3}$ $(-1/3+(-3))/(1+(-1/3)*(-3))=(-10/3)/2=-5/3$

Hence $(2 i - 6) \div (- 12 i + 4) = \frac{1}{2} (cos ρ + i sin ρ)$ $(2i-6)-:(-12i+4)=1/2(cosrho+isinrho)$ where $ρ = {tan}^{- 1} (- \frac{5}{3})$ $rho=tan^(-1)(-5/3)$

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How do you divide $\frac{2 i - 6}{- 12 i + 4}$ $( 2i-6) / ( -12i +4 )$ in trigonometric form?

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How do you divide $\frac{2 i - 6}{- 12 i + 4}$ $( 2i-6) / ( -12i +4 )$ in trigonometric form?