How do you divide $\frac{3 i + 8}{i + 4}$ $( 3i+8) / (i +4 )$ in trigonometric form?

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How do you divide $\frac{3 i + 8}{i + 4}$ $( 3i+8) / (i +4 )$ in trigonometric form?

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Answer 1 · 2016-05-21T09:49:03

$\frac{6 i - 4}{- 3 i - 5} = \sqrt{\frac{73}{17}} (cos ρ + i sin ρ)$ $(6i-4)/(-3i-5)=sqrt(73/17)(cosrho+isinrho)$ where $ρ = {tan}^{- 1} (\frac{4}{35})$ $rho=tan^(-1)(4/35)$

Explanation:

Let us first write $(3 i + 8)$ $(3i+8)$ and $(i + 4)$ $(i+4)$ 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 $3 i + 8 = (8 + 3 i) = \sqrt{8^{2} + 3^{2}} [cos α + i sin α]$ $3i+8=(8+3i)=sqrt(8^2+3^2)[cosalpha+isinalpha]$ or

$\sqrt{73} e^{i α}$ $sqrt73e^(ialpha)$ , where $tan α = \frac{3}{8}$ $tanalpha=3/8$ and

$i + 4 = (4 + i) = \sqrt{4^{2} + 1^{2}} [cos β + i sin β]$ $i+4=(4+i)=sqrt(4^2+1^2)[cosbeta+isinbeta]$ or

$\sqrt{17} e^{i β}$ $sqrt17e^(ibeta]$ , where $tan β = \frac{1}{4}$ $tanbeta=1/4$

Hence $\frac{3 i + 8}{i + 4} = \frac{\sqrt{73} e^{i α}}{\sqrt{17} e^{i β}} = \sqrt{\frac{73}{17}} e^{i (α - β)} = \sqrt{\frac{73}{17}} (cos (α - β) + i sin (α - β))$ $(3i+8)/(i+4)=(sqrt73e^(ialpha))/(sqrt17e^(ibeta])=sqrt(73/17)e^(i(alpha-beta))=sqrt(73/17)(cos(alpha-beta)+isin(alpha-beta))$

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

= $\frac{\frac{3}{8} - \frac{1}{4}}{1 + \frac{3}{8} \cdot \frac{1}{4}} = \frac{\frac{1}{8}}{1 + \frac{3}{32}} = \frac{1}{8} \cdot \frac{32}{35} = \frac{4}{35}$ $(3/8-1/4)/(1+3/8*1/4)=(1/8)/(1+3/32)=1/8*32/35=4/35$

Hence $\frac{6 i - 4}{- 3 i - 5} = \sqrt{\frac{73}{17}} (cos ρ + i sin ρ)$ $(6i-4)/(-3i-5)=sqrt(73/17)(cosrho+isinrho)$ where $ρ = {tan}^{- 1} (\frac{4}{35})$ $rho=tan^(-1)(4/35)$

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How do you divide $\frac{3 i + 8}{i + 4}$ $( 3i+8) / (i +4 )$ in trigonometric form?

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