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1 solution
Given that ( 1 2 − i 3 ) 5 + ( 1 2 − i 5 ) 5 = ( 1 2 + i ) 5 + ( 1 2 − I ) 5 = z 1 + w 1
where z 1 = ( 1 2 + i ) 5 and w 1 = ( 1 2 − i ) 5 .
Here z 1 = ( 1 2 + i ) 5 = ( ( 1 2 + i ) 2 ) 2 ( 1 2 + i ) = ( 1 4 3 + 2 4 i 2 ) 2 ( 1 2 + i ) = ( 1 9 8 7 3 + 6 8 6 4 i ) ( 1 2 + i ) = 2 3 1 6 1 2 + 1 0 2 2 4 1 i .
Similarly w 1 = ( 1 2 − i ) 5 = ( ( 1 2 − i ) 2 ) 2 ( 1 2 − i ) = ( 1 4 3 − 2 4 i 2 ) 2 ( 1 2 − i ) = ( 1 9 8 7 3 − 6 8 6 4 i ) ( 1 2 − i ) = 2 3 1 6 1 2 − 1 0 2 2 4 1 i .
Therefore, z 1 + w 1 = 2 3 1 6 1 2 + 1 0 2 2 4 i + 2 3 6 1 2 − 1 0 2 2 4 i ⟹ 4 6 3 2 2 4