3x^2+(1-i)x+3i=0

解

3 x^{2} + (1 - i) x + 3 i = 0

x = 0.55981 \dots - 0.55981 \dots i, x = - 0.89314 \dots + 0.89314 \dots i

解答ステップ

3 x^{2} + (1 - i) x + 3 i = 0

代用

x = a + bi

3 (a + bi)^{2} + (1 - i) (a + bi) + 3 i = 0

拡張

3 (a + bi)^{2} + (1 - i) (a + bi) + 3 i : (3 a^{2} - 3 b^{2} + a + b) + i (3 - a + b + 6 ab)

(3 a^{2} - 3 b^{2} + a + b) + i (3 - a + b + 6 ab) = 0

標準的な複素数形式で

0

を書き換える:

0 + 0 i

(3 a^{2} - 3 b^{2} + a + b) + i (3 - a + b + 6 ab) = 0 + 0 i

複素数は, その実数部と虚数部が等しい場合にのみ等しくなるequation系として書き換える:

[3 a^{2} - 3 b^{2} + a + b = 0 3 - a + b + 6 ab = 0]

[3 a^{2} - 3 b^{2} + a + b = 0 3 - a + b + 6 ab = 0] : (a = 0.55981 \dots, a = - 0.89314 \dots, b = - 0.55981 \dots b = 0.89314 \dots)

代用を戻す

x = a + bi

x = 0.55981 \dots - 0.55981 \dots i, x = - 0.89314 \dots + 0.89314 \dots i