(a+ib)^2=16+30i

解答

(a + ib)^{2} = 16 + 30 i

(a = 5, a = - 5, b = 3 b = - 3)

求解步骤

(a + ib)^{2} = 16 + 30 i

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

(a^{2} - b^{2}) + 2 iab = 16 + 30 i

复数仅在实部和虚部均相等时才相等改写为方程组:

[a^{2} - b^{2} = 16 2 ab = 30]

[a^{2} - b^{2} = 16 2 ab = 30] : (a = 5, a = - 5, b = 3 b = - 3)

(a = 5, a = - 5, b = 3 b = - 3)

z^{2} - i \cdot z + 2 = 0

\frac{3 - a - 2 b}{5} - 3 i + a + bi = 4 - i

x^{2} = - 1 + 3 i

x + 3 y i = 4 - 7 i

(5 + 9 i)^{2} = a + bi