How do you divide $( -i+8) / (2i +7 )$ in trigonometric form?

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Answer 1 · 2018-07-11T14:50:54

$1/53(54-23i)$

We have

$\frac{-i+8}{2i+7}$

$=\frac{8-i}{7+2i}$

$=\frac{\sqrt{65}(\cos(-\tan^{-1}(1/8))+i\sin(-\tan^{-1}(1/8)))}{\sqrt{53}(\cos(\tan^{-1}(2/7))+i\sin(\tan^{-1}(2/7)))}$

$=\sqrt{\frac{65}{53}}(\cos(-\tan^{-1}(1/8)-\tan^{-1}(2/7))+i\sin(-\tan^{-1}(1/8)-\tan^{-1}(2/7)))$

$=\sqrt{\frac{65}{53}}(\cos(-\tan^{-1}(23/54))+i\sin(-\tan^{-1}(23/54)))$

$=\sqrt{\frac{65}{53}}(54/\sqrt3445-i23/\sqrt{3445})$

$=\sqrt{\frac{65}{53\times 3445}}(54-23i)$

$=1/53(54-23i)$

How do you divide ( -i+8) / (2i +7 )( -i+8) / (2i +7 ) in trigonometric form?