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Question
Show that `(2 + "i"sqrt(3))^10 - (2 - "i" sqrt(3))^10` is purely imaginary
Solution
`(20 - 5"i")/(7 - 6"i") = (20 - 5"i")/(7 - 6"i") xx (7 + 6"i")/(7 + 6"i")`
= `(140 + 120"I" - 35"I" + 30)/(49 + 36)`
= `(10 + 85"I")/85`
= `(85(2 + "I"))/85`
= (2 + I)
z = `((19 - 7"i")/(9 + "i"))^12 + ((20 - 5"i")/(7 - 6"i"))^12`
= `(2 - "i")^2 + (2 + "i")^12`
`bar(z) = bar((2 - "i")^12 + (2 + "i")^12)`
= `bar((2 - "i")^12) + bar((2 + "i"))^12`
= `(2 + "i")^12 + (2 - "i")^12`
`bar(z)` = z
⇒ z is purely real.
Let z = `(2 + "i" sqrt(3))^10 - (2 - "i" sqrt(3))^10`
`bar(z) = bar((2 + "i" sqrt(3))^10) - bar((2 - "i" sqrt(3))^10)`
= `bar((2 + "i" sqrt(3))^10 - (2 - "i" sqrt(3))^10)`
= `(2 - "i" sqrt(3))^10 - (2 + "i" sqrt(3))^10`
= `[(2 + "i" sqrt(3))^10 - (2 - "i" sqrt(3))^10]`
`bar(z)` = z
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