How do you divide $\frac{- 4 i + 9}{2 i - 12}$ $( -4i+9) / (2i-12)$ in trigonometric form?

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How do you divide $\frac{- 4 i + 9}{2 i - 12}$ $( -4i+9) / (2i-12)$ in trigonometric form?

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Answer 1 · 2016-02-06T15:33:41

$\frac{58 - 3 i}{74}$ $(58-3i)/74$ More calculations are given below.

Explanation:

Multiply the denominator, with its conjugate. It is 2i+12 here. Now multiplication would work as follows:

$\frac{(- 4 i + 9) (2 i + 12)}{(- 2 i + 12) (2 i + 12)}$ $((-4i+9) ( 2i+12))/ ((-2i+12) (2i+12))$

= $\frac{8 + 108 - 24 i + 18 i}{4 + 144}$ $(8 +108 -24i+18i)/(4+144)$

= $\frac{116 - 6 i}{148}$ $(116-6i)/148$

= $\frac{58 - 3 i}{74}$ $(58-3i)/74$

= $\frac{58}{74} - i \frac{3}{74}$ $58/74 - i 3/74$

Now, let $r cos θ = \frac{58}{74}$ $r cos theta = 58/74$ and $r sin θ = \frac{3}{74}$ $r sin theta= 3/74$

On squaring and adding $r^{2} = \frac{3373}{5476}$ $r^2=3373/5476$ that means $r = \frac{\sqrt{3373}}{74}$ $r=sqrt3373/74$ and on division it would be $tan θ = - \frac{3}{58}$ $tan theta=-3/58$ , $θ = {tan}^{- 1} (- \frac{3}{58})$ $theta= tan^-1 (-3/58)$

The required trignometric form would be $r (cos θ + i sin θ)$ $r(cos theta+isin theta)$ , where r and $θ$ $theta$ would have values as worked out above.

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How do you divide $\frac{- 4 i + 9}{2 i - 12}$ $( -4i+9) / (2i-12)$ in trigonometric form?

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Explanation:

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How do you divide $\frac{- 4 i + 9}{2 i - 12}$ $( -4i+9) / (2i-12)$ in trigonometric form?