What is the answer to i(t)=t^2-2t+1 ?

The answer to i(t)=t^2-2t+1 is t=0.37518…-0.30024…i,t=1.62481…+1.30024…i

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

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

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

Solution steps

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

Substitute

t = x + y i

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

Expand

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

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[- 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)

Substitute back

t = x + y i

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}

What is the answer to i(t)=t^2-2t+1 ?
The answer to i(t)=t^2-2t+1 is t=0.37518…-0.30024…i,t=1.62481…+1.30024…i