solvefor z,z^2=(x+iy)^2

解答

求解

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

z = x + y i, z = - x - y i, z = - y^{2} + \frac{x y}{- y ^{2}} i, z = - - y^{2} - \frac{x y}{- y ^{2}} i

求解步骤

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

替代

z = x + y i

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

乘开

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

复数仅在实部和虚部均相等时才相等改写为方程组:

[x^{2} - y^{2} = x^{2} - y^{2} 2 x y = 2 x y]

[x^{2} - y^{2} = x^{2} - y^{2} 2 x y = 2 x y] : x = x, x = - x, x = - y^{2}, x = - - y^{2}, y = y y = - y y = \frac{x y}{- y ^{2}} y = - \frac{x y}{- y ^{2}}

z = x + y i

代回

z = x + y i, z = - x - y i, z = - y^{2} + \frac{x y}{- y ^{2}} i, z = - - y^{2} - \frac{x y}{- y ^{2}} i

∣ 2 i + 6 j + 2 ik - 4 jk ∣ = 5

(x + y i) + (4 - 7 i) = 3 - 4 i

z^{2} = 5 + 12 i

z^{2} - 4 z i + 4 i = 7

(1 + i) x - i y = 2 + i