How do you divide #( 8i+2) / (-i +5)# in trigonometric form?

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How do you divide #( 8i+2) / (-i +5)# in trigonometric form?

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Answer 1 · 2016-01-28T07:16:47

#1/13+21/13i#

Explanation:

#(8i+2)/(-i+5)#
#=(2+8i)/(5-i)#
#=((2+8i)(5+i))/((5-i)(5+i))#
#=(10+40i+2i+8i^2)/(25-i^2)#
#=(10+42i+8(-1))/(25-(-1))# [as, #i^2=(root2 (-1))^2=-1#]
#=(10+42i-8)/(25+1)#
#=(2+42i)/26#
#=(2(1+21i))/(2.13)#
#=(1+21i)/13#
#=1/13+21/13i#

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How do you divide #( 8i+2) / (-i +5)# in trigonometric form?

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