(a+bi)^2=15+8i

解答

(a + bi)^{2} = 15 + 8 i

(a = 4, a = - 4, b = 1 b = - 1)

求解步骤

(a + bi)^{2} = 15 + 8 i

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

(a^{2} - b^{2}) + 2 iab = 15 + 8 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} = 15 2 ab = 8]

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

(a = 4, a = - 4, b = 1 b = - 1)

i = - 0.3 q^{2} + 30 q

so l v e f or x, s i x = x (x + 1) - x (x - 1)

(z - (2 + i)) (z - (2 - i)) = 0

(1 + q) (2 + i) + p = ((p + q - 1) (3 - i) + 2)

10 x - 6 i y = 700