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

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

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Answer 1 · 2016-07-25T11:08:48

In trigonometric form $Z = \sqrt{2} [cos 188.13 + i sin 188.13]$ $Z=sqrt2[cos188.13+isin188.13]$

Explanation:

Let $Z = \frac{2 i - 6}{- 2 i + 4} = \frac{(2 i - 6) (4 + 2 i)}{(4 + 2 i) (4 - 2 i)} = \frac{4 i^{2} - 4 i - 24}{16 - 4 i^{2}} = \frac{- 4 i - 28}{20} = - \frac{7}{5} - \frac{1}{5} i$ $Z=(2i-6)/(-2i+4)= ((2i-6)(4+2i))/((4+2i)(4-2i))= (4i^2-4i-24)/(16-4i^2) =(-4i-28)/20= -7/5-1/5i$ [since $i^{2} = - 1$ $i^2=-1$ ] Modulus $Z = \sqrt{{(- \frac{7}{5})}^{2} + {(- \frac{1}{5})}^{2}} = \sqrt{2}$ $Z=sqrt((-7/5)^2+(-1/5)^2) =sqrt2$ Argument Z: $θ = {tan}^{- 1} (\frac{1}{5} / \frac{7}{5}) = {tan}^{- 1} (\frac{1}{7}) = {8.13}^{0} + 180^{0} = {188.13}^{0} [180^{0}$ $theta=tan^-1(1/5/7/5)=tan^-1(1/7)=8.13^0+180^0=188.13^0[180^0$ is added as it is on 3rd quadrant] Hence in trigonometric form $Z = \sqrt{2} [cos 188.13 + i sin 188.13]$ $Z=sqrt2[cos188.13+isin188.13]$ [Ans]

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

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How do you divide 2i−6−2i+4( 2i-6) / ( -2i +4 ) in trigonometric form?

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