a-bi(a+bi+1)=11+3i

解答

a - bi (a + bi + 1) = 11 + 3 i

a = 10.93683 \dots, a = - 0.09953 \dots, a = - 1.83730 \dots, b = - 0.25132 \dots b = - 3.33159 \dots b = 3.58291 \dots

求解步骤

a - bi (a + bi + 1) = 11 + 3 i

展开

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

(a + b^{2}) + i (- b - ab) = 11 + 3 i

复数仅在实部和虚部均相等时才相等改写为方程组:

[a + b^{2} = 11 - b - ab = 3]

[a + b^{2} = 11 - b - ab = 3] : a = 10.93683 \dots, a = - 0.09953 \dots, a = - 1.83730 \dots, b = - 0.25132 \dots b = - 3.33159 \dots b = 3.58291 \dots

a = 10.93683 \dots, a = - 0.09953 \dots, a = - 1.83730 \dots, b = - 0.25132 \dots b = - 3.33159 \dots b = 3.58291 \dots

\frac{360}{i r} - 1500 = 0

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

1 0^{3} = \frac{i}{i o}

a + ib = 3 + 2 i

- 0.008647 i = \frac{1}{x \cdot i}