How do you divide $\frac{2 i - 1}{3 i + 5}$ $( 2i-1) / ( 3i +5 )$ in trigonometric form?

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How do you divide $\frac{2 i - 1}{3 i + 5}$ $( 2i-1) / ( 3i +5 )$ in trigonometric form?

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Answer 1 · 2016-05-21T09:56:19

$\frac{2 i - 1}{3 i + 5} = \sqrt{\frac{5}{34}} (cos ρ + i sin ρ)$ $(2i-1)/(3i+5)=sqrt(5/34)(cosrho+isinrho)$ where $ρ = {tan}^{- 1} 13$ $rho=tan^(-1)13$

Explanation:

Let us first write $(2 i - 1)$ $(2i-1)$ and $(3 i + 5)$ $(3i+5)$ 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 $2 i - 1 = (- 1 + 2 i) = \sqrt{{(- 1)}^{2} + 2^{2}} [cos α + i sin α]$ $2i-1=(-1+2i)=sqrt((-1)^2+2^2)[cosalpha+isinalpha]$ or

$\sqrt{5} e^{i α}$ $sqrt5e^(ialpha)$ , where $tan α = \frac{6}{- 4} = \frac{2}{- 1} = - 2$ $tanalpha=6/(-4)=2/(-1)=-2$ and

$3 i + 5 = (5 + 3 i) = \sqrt{5^{2} + 3^{2}} [cos β + i sin β]$ $3i+5=(5+3i)=sqrt(5^2+3^2)[cosbeta+isinbeta]$ or

$\sqrt{34} e^{i β}$ $sqrt34e^(ibeta]$ , where $tan β = \frac{3}{5}$ $tanbeta=3/5$

Hence $\frac{2 i - 1}{3 i + 5} = \frac{\sqrt{5} e^{i α}}{\sqrt{34} e^{i β}} = \sqrt{\frac{5}{34}} e^{i (α - β)} = \sqrt{\frac{5}{34}} (cos (α - β) + i sin (α - β))$ $(2i-1)/(3i+5)=(sqrt5e^(ialpha))/(sqrt34e^(ibeta])=sqrt(5/34)e^(i(alpha-beta))=sqrt(5/34)(cos(alpha-beta)+isin(alpha-beta))$

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

= $\frac{- 2 - \frac{3}{5}}{1 + (- 2) \cdot (\frac{3}{5})}$ $(-2-3/5)/(1+(-2)*(3/5))$ = $\frac{- \frac{13}{5}}{- \frac{1}{5}} = - \frac{13}{5} \cdot \frac{- 5}{1} = 13$ $(-13/5)/(-1/5)=-13/5*(-5)/1=13$

Hence $\frac{2 i - 1}{3 i + 5} = \sqrt{\frac{5}{34}} (cos ρ + i sin ρ)$ $(2i-1)/(3i+5)=sqrt(5/34)(cosrho+isinrho)$ where $ρ = {tan}^{- 1} 13$ $rho=tan^(-1)13$

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How do you divide $\frac{2 i - 1}{3 i + 5}$ $( 2i-1) / ( 3i +5 )$ in trigonometric form?

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