Question

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

Answer 1

`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`