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

解答

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

t = 0.37518 \dots - 0.30024 \dots i, t = 1.62481 \dots + 1.30024 \dots i

求解步骤

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

替代

t = x + y i

i (x + y i) = (x + y i)^{2} - 2 (x + y i) + 1

乘开

- y + i x = (x^{2} - y^{2} - 2 x + 1) + i (- 2 y + 2 x y)

复数仅在实部和虚部均相等时才相等改写为方程组:

[- y = x^{2} - y^{2} - 2 x + 1 x = - 2 y + 2 x y]

[- y = x^{2} - y^{2} - 2 x + 1 x = - 2 y + 2 x y] : (x = 0.37518 \dots, x = 1.62481 \dots, y = - 0.30024 \dots y = 1.30024 \dots)

t = x + y i

代回

t = 0.37518 \dots - 0.30024 \dots i, t = 1.62481 \dots + 1.30024 \dots i

\frac{Z}{1 + Z} = 3 + i 4

x + 2 y i = (1 + 3 i) (1 + i)

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

2 x + 7 + (3 - y) i = - 4 + 6 i

a + bi = (a - bi)^{2}