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)#