7r=3t^2i-2j

解答

7 r = 3 t^{2} i - 2 j

(r = - \frac{2 j}{7}, t = 0)

求解步骤

7 r = 3 t^{2} i - 2 j

将

7 r

改写成标准复数形式:

7 r + 0 i

7 r + 0 i = 3 t^{2} i - 2 j

将

3 t^{2} i - 2 j

改写成标准复数形式:

- 2 j + 3 t^{2} i

7 r + 0 i = - 2 j + 3 t^{2} i

复数仅在实部和虚部均相等时才相等改写为方程组:

[7 r = - 2 j 0 = 3 t^{2}]

[7 r = - 2 j 0 = 3 t^{2}] : (r = - \frac{2 j}{7}, t = 0)

(r = - \frac{2 j}{7}, t = 0)

i z^{2} - (2 + 3 i) z + 1 + 5 i = 0

z^{2} (3 - i) + 5 (1 - i) (1 + i) + 2 + 16 i = 0

so l v e f or y, s i x^{2} + y^{2} = z^{2}

\frac{( 2 - 3 i )}{2 x + y i} = 3 + 2 i

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