a-bi-1=(15+8i)/(4(a+bi))

解

a - bi - 1 = \frac{15 + 8 i}{4 ( a + bi )}

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

解答ステップ

a - bi - 1 = \frac{15 + 8 i}{4 ( a + bi )}

以下で両辺を乗じる:

4 a + 4 ib

a (4 a + 4 ib) - bi (4 a + 4 ib) - 1 \cdot (4 a + 4 ib) = \frac{15 + 8 i}{4 ( a + bi )} (4 a + 4 ib)

拡張

(4 a^{2} - 4 a + 4 b^{2}) - 4 ib = 15 + 8 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:

[4 a^{2} - 4 a + 4 b^{2} = 15 - 4 b = 8]

[4 a^{2} - 4 a + 4 b^{2} = 15 - 4 b = 8] : (a = \frac{1}{2}, b = - 2)

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