r(t)=(2t+1)i+(t^2-1)j

解答

r (t) = (2 t + 1) i + (t^{2} - 1) j

(r = \frac{3}{2} j, t = - \frac{1}{2})

求解步骤

r (t) = (2 t + 1) i + (t^{2} - 1) j

将

r (t)

改写成标准复数形式:

r (t) + 0 i

r (t) + 0 i = (2 t + 1) i + (t^{2} - 1) j

将

(2 t + 1) i + (t^{2} - 1) j

改写成标准复数形式:

(j t^{2} - j) + (2 t + 1) i

r (t) + 0 i = (j t^{2} - j) + (2 t + 1) i

复数仅在实部和虚部均相等时才相等改写为方程组:

[r (t) = j t^{2} - j 0 = 2 t + 1]

[r (t) = j t^{2} - j 0 = 2 t + 1] : (r = \frac{3}{2} j, t = - \frac{1}{2})

(r = \frac{3}{2} j, t = - \frac{1}{2})

2 x + 51 = 4 (2 + y i) - 12

z^{2} - (1 + 9 i) z - 20 + 5 i = 0

z = z (a + i)

\frac{x + 1 + i ( y - 3 )}{5 + 3 i} = 1 + i

v \cdot v = i - 5 j