k_{1}-3-2sqrt(7)ik_{2}=0

解答

k_{1} - 3 - 27 i k_{2} = 0

k_{1} = 3, k_{2} = 0

求解步骤

k_{1} - 3 - 27 i k_{2} = 0

将

k_{1} - 3 - 27 i k_{2}

改写成标准复数形式:

(k_{1} - 3) - 27 k_{2} i

(k_{1} - 3) - 27 k_{2} i = 0

将

0

改写成标准复数形式:

0 + 0 i

(k_{1} - 3) - 27 k_{2} i = 0 + 0 i

复数仅在实部和虚部均相等时才相等改写为方程组:

[k_{1} - 3 = 0 - 27 k_{2} = 0]

[k_{1} - 3 = 0 - 27 k_{2} = 0] : k_{1} = 3, k_{2} = 0

k_{1} = 3, k_{2} = 0

so l v e f or c, i = \frac{d}{( 1 - c )}

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

\frac{3 + ai}{5 + bi} = 3 + 5 i

so l v e f or x, (x + y)^{2} = 3 - 4 i

z^{2} = 3 + 3 i