(1+ai)/(6+bi)=1+5i

解答

\frac{1 + ai}{6 + bi} = 1 + 5 i

a = 31, b = 1

求解步骤

\frac{1 + ai}{6 + bi} = 1 + 5 i

在两边乘以

6 + bi

\frac{1 + ai}{6 + bi} (6 + bi) = 1 \cdot (6 + bi) + 5 i (6 + bi)

化简

1 + ia = (6 - 5 b) + i (30 + b)

复数仅在实部和虚部均相等时才相等改写为方程组:

[1 = 6 - 5 b a = 30 + b]

[1 = 6 - 5 b a = 30 + b] : a = 31, b = 1

a = 31, b = 1

0 = \frac{1 + λi}{1 - λi}

z^{2} - (1 - 2 i) z + (2 - 16 i) = 0

(8 + 9 i) y - (3 + 6 i) x = 16 - 3 i

(5 - 3 i) y - (6 + 2 i) x = 3 - 41 i

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