How do you divide # (-3+5i)/(2-i) # in trigonometric form?

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How do you divide # (-3+5i)/(2-i) # in trigonometric form?

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Answer 1 · 2017-08-15T14:39:33

#(-3+5i)/(2-i) = 1/5sqrt (170)(cos2.57+isin2.57)#

Explanation:

#(-3+5i)/(2-i)=((-3+5i)(2+i))/((2-i)(2+i))= (-11+7i)/5 = -11/5+7/5i#

Use mod-arg form, #a+bi=r(cosvartheta+isinvartheta)# where #r= sqrt (-11/5^2+7/5^2) = sqrt (34/5) = 1/5sqrt170# and #vartheta = arctan ((7/5)/(-11/5))+pi = 2.57^"c"#

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How do you divide # (-3+5i)/(2-i) # in trigonometric form?

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