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Question
Answer the following:
Simplify: `("i"^29 + "i"^39 + "i"^49)/("i"^30 + "i"^40 + "i"^50)`
Solution
`("i"^29 + "i"^39 + "i"^49)/("i"^30 + "i"^40 + "i"^50)`
= `("i"^29(1 + "i"^10 + "i"^20))/("i"^30(1 + "i"^10 + "i"^20)`
= `1/"i" = 1/"i" xx "i"/"i" = "i"/"i"^2`
= `"i"/(-1)`
= – i.
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