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

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Answer 1 · 2018-07-11T16:19:39

#1/25(11+27i)#

We have

#\frac{2+8i}{7+i}#

#=\frac{\sqrt{68}(\cos(\tan^{-1}(4))+i\sin(\tan^{-1}(4)))}{\sqrt{50}(\cos(\tan^{-1}(1/7))+i\sin(\tan^{-1}(1/7)))}#

#=\sqrt{\frac{68}{50}}(\cos(\tan^{-1}(4)-\tan^{-1}(1/7))+i\sin(\tan^{-1}(4)-\tan^{-1}(1/7)))#

#=\sqrt{\frac{34}{25}}(\cos(\tan^{-1}(27/11))+i\sin(\tan^{-1}(27/11)))#

#=\sqrt{\frac{34}{25}}(11/\sqrt850+i27/\sqrt{850})#

#=\sqrt{\frac{34}{25\times 850}}(11+27i)#

#=1/25(11+27i)#