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प्रश्न
Show that `((19 - 7"i")/(9 + "i"))^12 + ((20 - 5"i")/(7 - 6"i"))^12` is real
उत्तर
`(19 - 7"i")/(9 + "i") xx (9 - "i")/(9 - "i") = (164 - 82"i")/82`
= 2 – i
`(20 - 5"i")/(7 - 6"i") xx (7 + 6"i")/( + 6"i") = (170 + 85"i")/85`
= 2 + i
∴ `((19 - 7"i")/(9 + "i"))^12 + ((20 - 5"i")/(7 - 6"i"))^12`
= `(2 - "i")^12 + (12 + "i")^12`
Let z = `(2 - "i")^12 + (2 + "i")^12`
Then `bar(z) = bar((2 - "i")^12 + (2 + "i")^12` ......`[because bar(z_1 + z_2) = bar(z_1) + bar(z_2)]`
= `bar((2 - 1)^12) + bar((2 + "i")^12)` ......`[because bar((z^"n")) = bar((z))^"n"]`
= `(2 + "i")^12 + (2 - "i")^12`
∴ `bar(z)` = z
∴ z is real.
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