solvefor a,z=(1-ai)/(1+ai)

解答

求解

a, z = \frac{1 - ai}{1 + ai}

a = - \frac{1 - z}{1 + z} i

求解步骤

z = \frac{1 - ai}{1 + ai}

替代

a = x + y i

z = \frac{1 - ( x + y i ) i}{1 + ( x + y i ) i}

在两边乘以

1 + (x + y i) i

z (1 + (x + y i) i) = \frac{1 - ( x + y i ) i}{1 + ( x + y i ) i} (1 + (x + y i) i)

乘开

z - zy + i z x = (1 + y) - i x

复数仅在实部和虚部均相等时才相等改写为方程组:

[z - zy = 1 + y z x = - x]

[z - zy = 1 + y z x = - x] : x = 0, y = - \frac{1 - z}{1 + z}, z \neq = - 1

a = x + y i

代回

a = - \frac{1 - z}{1 + z} i

(1 - i) x + (3 + i) y = - 3 + 3 i

(2 + i) x + 7 y = 1 + 2 i, (1 - i) x - i y = 4 - i

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

i (x) = \frac{x - 1}{x + 3}

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