展開する (x-i)^2(x-1+i)^2

解

展開する

(x - i)^{2} (x - 1 + i)^{2}

(x^{4} - 2 x^{3} - 2 x + 3 x^{2}) + (2 + 2 x^{2} - 2 x) i

解答ステップ

(x - i)^{2} (x - 1 + i)^{2}

拡張

(x - i)^{2} : x^{2} - 2 x i - 1

= (x^{2} - 2 x i - 1) (x - 1 + i)^{2}

= (x - 1 + i)^{2} (x^{2} - 2 x i - 1)

分配法則を適用する:

m (a + b + c) = ma + mb + m c

(x - 1 + i)^{2} (x^{2} - 2 x i - 1) = (x - 1 + i)^{2} x^{2} + (x - 1 + i)^{2} (- 2 x i) + (x - 1 + i)^{2} (- 1)

= (x - 1 + i)^{2} x^{2} + (x - 1 + i)^{2} (- 2 x i) + (x - 1 + i)^{2} (- 1)

簡素化

(x - 1 + i)^{2} x^{2} + (x - 1 + i)^{2} (- 2 x i) + (x - 1 + i)^{2} (- 1) : 2 i + x^{4} - 2 x^{3} + 2 x^{2} i - 2 x + 3 x^{2} - 2 x i

= 2 i + x^{4} - 2 x^{3} + 2 x^{2} i - 2 x + 3 x^{2} - 2 x i

標準的な複素数形式で書き直す:

(x^{4} - 2 x^{3} - 2 x + 3 x^{2}) + (2 + 2 x^{2} - 2 x) i

= (x^{4} - 2 x^{3} - 2 x + 3 x^{2}) + (2 + 2 x^{2} - 2 x) i