(1+sqrt(3)i)^{11}=a+ib

解答

(1 + 3 i)^{11} = a + ib

a = 1024, b = - 10243

求解步骤

(1 + 3 i)^{11} = a + ib

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(1 + 3 i)^{11} : 1024 - 10243 i

1024 - 10243 i = a + ib

复数仅在实部和虚部均相等时才相等改写为方程组:

[1024 = a - 10243 = b]

[1024 = a - 10243 = b] : a = 1024, b = - 10243

a = 1024, b = - 10243

∣ z - i ∣ = 5

z = 6 e^{\frac{π}{3} i}

z^{2} - i = 0

(3 x + 2 y i)^{2} = \frac{200}{4 + 3 i}

2 (5 x + 9 y) - (9 y - 16 x) i = - 11 - 5 i