Question

Show that `((19 - 7"i")/(9 + "i"))^12 + ((20 - 5"i")/(7 - 6"i"))^12` is real

Answer 1

`(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.