a+bi+1/(a+bi)=1

解

a + bi + \frac{1}{a + bi} = 1

(a = \frac{1}{2}, a = \frac{1}{2}, b = \frac{3}{2} b = - \frac{3}{2})

解答ステップ

a + bi + \frac{1}{a + bi} = 1

以下で両辺を乗じる:

a + bi

a (a + bi) + bi (a + bi) + \frac{1}{a + bi} (a + bi) = 1 \cdot (a + bi)

拡張

(a^{2} - b^{2} + 1) + 2 iab = a + ib

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:

[a^{2} - b^{2} + 1 = a 2 ab = b]

[a^{2} - b^{2} + 1 = a 2 ab = b] : (a = \frac{1}{2}, a = \frac{1}{2}, b = \frac{3}{2} b = - \frac{3}{2})

(a = \frac{1}{2}, a = \frac{1}{2}, b = \frac{3}{2} b = - \frac{3}{2})