T ı ˊ nh ( 3 ​ − i ) 4 ( 2 i ) 5

Giải thích

Tính ( 3 ​ − i ) 4 ( 2 i ) 5 - Đặt x = ( 2 i ) 5 ta có x = 2 5 ( i ) 4 i = 2 5 . i = 2 5 ( 1 + i ; 2 5 ( cos 9 0 0 + i sin 9 0 0 ) ( 1 ) - Đặt y = ( 3 ​ − i ) 4 ta có ( 3 ​ − i ) = 2 ( 2 3 ​ ​ − 2 i ​ ) = 2 ( cos ( − 3 0 ∘ ) + i sin ( − 3 0 ∘ ) ) ⇒ ( 3 ​ − i ) 4 = 2 4 ( cos ( − 4.3 0 ∘ ) + i sin ( − 4.3 0 ∘ ) ) ​ hay y = 2 4 ( cos ( − 12 0 ∘ ) + i sin ( − 12 0 ∘ ) ) ( 2 ) Từ (1), (2)suy ra x y = 2 5 + 4 [ ( cos ( 9 0 ∘ − 12 0 ∘ ) + i sin ( 9 0 ∘ − 12 0 ∘ ) ] = 2 9 [ ( cos ( − 3 0 0 ) + i sin ( − 3 0 0 ) ] = 2 9 ( cos 3 0 0 − i sin 3 0 0 ) = 2 9 ( 2 3 ​ ​ − 2 i ​ ) = 2 8 ( 3 ​ − i ) . Vậy ( 3 ​ − i ) 4 ( 2 i ) 5 = 2 8 ( 3 ​ − i )