How do you divide $\frac{8 i + 2}{- i + 5}$ $( 8i+2) / (-i +5)$ in trigonometric form?

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How do you divide $\frac{8 i + 2}{- i + 5}$ $( 8i+2) / (-i +5)$ in trigonometric form?

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Answer 1 · 2016-01-28T07:16:47

$\frac{1}{13} + \frac{21}{13} i$ $1/13+21/13i$

Explanation:

$\frac{8 i + 2}{- i + 5}$ $(8i+2)/(-i+5)$
$= \frac{2 + 8 i}{5 - i}$ $=(2+8i)/(5-i)$
$= \frac{(2 + 8 i) (5 + i)}{(5 - i) (5 + i)}$ $=((2+8i)(5+i))/((5-i)(5+i))$
$= \frac{10 + 40 i + 2 i + 8 i^{2}}{25 - i^{2}}$ $=(10+40i+2i+8i^2)/(25-i^2)$
$= \frac{10 + 42 i + 8 (- 1)}{25 - (- 1)}$ $=(10+42i+8(-1))/(25-(-1))$ [as, $i^{2} = {(\sqrt[2]{- 1})}^{2} = - 1$ $i^2=(root2 (-1))^2=-1$ ]
$= \frac{10 + 42 i - 8}{25 + 1}$ $=(10+42i-8)/(25+1)$
$= \frac{2 + 42 i}{26}$ $=(2+42i)/26$
$= \frac{2 (1 + 21 i)}{2.13}$ $=(2(1+21i))/(2.13)$
$= \frac{1 + 21 i}{13}$ $=(1+21i)/13$
$= \frac{1}{13} + \frac{21}{13} i$ $=1/13+21/13i$

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How do you divide $\frac{8 i + 2}{- i + 5}$ $( 8i+2) / (-i +5)$ in trigonometric form?

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Explanation:

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How do you divide $\frac{8 i + 2}{- i + 5}$ $( 8i+2) / (-i +5)$ in trigonometric form?