How do you divide $\frac{- 6 i + 5}{i + 2}$ $( - 6 i + 5) / (i+ 2 )$ in trigonometric form?

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How do you divide $\frac{- 6 i + 5}{i + 2}$ $( - 6 i + 5) / (i+ 2 )$ in trigonometric form?

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Answer 1 · 2018-08-09T09:39:03

$\frac{- 6 i + 5}{i + 2} = \sqrt{\frac{61}{5}} c i s (- {tan}^{- 1} (\frac{17}{- 9}))$ $(-6i+5)/(i+2)=sqrt(61/5)cis(-tan^-1(17/-9))$

Explanation:

$\frac{- 6 i + 5}{i + 2} = f (r, θ)$ $(-6i+5)/(i+2)=f(r,theta)$
$z_{1} = - 6 i + 5$ $z_1 = -6i+5$
$r_{1} = \sqrt{{(- 6)}^{2} + 5^{2}} = \sqrt{36 + 25} = \sqrt{61}$ $r_1 = sqrt((-6)^2+5^2) =sqrt(36+25)=sqrt61$
$θ_{1} = {tan}^{- 1} (\frac{5}{- 6}) = 2 π - {tan}^{- 1} (\frac{5}{6})$ $theta_1=tan^-1 (5/-6)=2pi-tan^-1(5/6)$
$z_{2} = i + 2$ $z_2=i+2$
$r_{2} = \sqrt{1^{2} + 2^{2}} = \sqrt{1 + 4} = \sqrt{5}$ $r_2 = sqrt(1^2+2^2) =sqrt(1+4)=sqrt5$
$θ_{2} = {tan}^{- 1} (\frac{2}{1}) = 2 π + {tan}^{- 1} 2$ $theta_2=tan^-1 (2/1)=2pi+tan^-1 2$
Thus,
$\frac{- 6 i + 5}{i + 2} = \frac{\sqrt{61}, (2 π - {tan}^{- 1} (\frac{5}{6}))}{\sqrt{5}, (2 π + {tan}^{- 1} 2)}$ $(-6i+5)/(i+2)=(sqrt61, (2pi-tan^-1(5/6)))/(sqrt5, (2pi+tan^-1 2))$
By De-Moivre's Theorem,

$\frac{\sqrt{61}, (2 π - {tan}^{- 1} (\frac{5}{6}))}{\sqrt{5}, (2 π + {tan}^{- 1} 2)}$ $(sqrt61, (2pi-tan^-1(5/6)))/(sqrt5, (2pi+tan^-1 2))$
$= \frac{\sqrt{61}}{\sqrt{5}} \times c i s (2 π - {tan}^{- 1} (\frac{5}{6}) - (2 π + {tan}^{- 1} 2))$ $=sqrt 61 /(sqrt 5) xxcis(2pi-tan^-1(5/6)-(2pi+tan^-1 2))$

$= \sqrt{\frac{61}{5}} c i s (- {tan}^{- 1} (\frac{5}{6}) - {tan}^{- 1} 2)$ $=sqrt(61/5)cis(-tan^-1(5/6)-tan^-1 2)$
$= \sqrt{\frac{61}{5}} c i s (- ({tan}^{- 1} (\frac{5}{6}) + {tan}^{- 1} 2))$ $=sqrt(61/5)cis(-(tan^-1(5/6)+tan^-1 2))$

${tan}^{- 1} (\frac{5}{6}) + {tan}^{- 1} 2 = {tan}^{- 1} (\frac{\frac{5}{6} + 2}{1 - \frac{5}{6} \times 2})$ $tan^-1 (5/6) +tan^-1 2 = tan^-1 ((5/6+2)/(1-5/6xx2))$

$= {tan}^{- 1} (\frac{5 \times 1 + 2 \times 6}{1 \times 6 - 5 \times 3}) = {tan}^{- 1} (\frac{5 + 12}{6 - 15})$ $=tan^-1((5xx1+2xx6)/(1xx6-5xx3))=tan^-1 ((5+12)/(6-15))$

$= {tan}^{- 1} (\frac{17}{- 9})$ $=tan^-1(17/-9)$
Thus,
$\frac{- 6 i + 5}{i + 2} = \sqrt{\frac{61}{5}} c i s (- {tan}^{- 1} (\frac{17}{- 9}))$ $(-6i+5)/(i+2)=sqrt(61/5)cis(-tan^-1(17/-9))$

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How do you divide $\frac{- 6 i + 5}{i + 2}$ $( - 6 i + 5) / (i+ 2 )$ in trigonometric form?

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Science

Math

Humanities

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How do you divide −6i+5i+2( - 6 i + 5) / (i+ 2 ) in trigonometric form?

1 Answer

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How do you divide $\frac{- 6 i + 5}{i + 2}$ $( - 6 i + 5) / (i+ 2 )$ in trigonometric form?