What is the answer to z^2-2z+1-i=0 ?

The answer to z^2-2z+1-i=0 is z=(-sqrt(2)+2)/2-1/(sqrt(2))i,z=(sqrt(2)+2)/2+1/(sqrt(2))i

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z^2-2z+1-i=0

z^{2} - 2 z + 1 - i = 0

z = \frac{- 2 + 2}{2} - \frac{1}{2} i, z = \frac{2 + 2}{2} + \frac{1}{2} i

Solution steps

z^{2} - 2 z + 1 - i = 0

Substitute

z = x + y i

(x + y i)^{2} - 2 (x + y i) + 1 - i = 0

Expand

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

(x^{2} - y^{2} - 2 x + 1) + i (- 1 - 2 y + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 2 x + 1) + i (- 1 - 2 y + 2 x y) = 0 + 0 i

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

[x^{2} - y^{2} - 2 x + 1 = 0 - 1 - 2 y + 2 x y = 0]

[x^{2} - y^{2} - 2 x + 1 = 0 - 1 - 2 y + 2 x y = 0] : (x = \frac{- 2 + 2}{2}, x = \frac{2 + 2}{2}, y = - \frac{1}{2} y = \frac{1}{2})

Substitute back

z = x + y i

z = \frac{- 2 + 2}{2} - \frac{1}{2} i, z = \frac{2 + 2}{2} + \frac{1}{2} i

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

z^{2} - z (1 + 4 i) + 3 + 11 i = 0

x^{2} - 6 i^{2} = 0

2 x^{2} + 9 x - i = 0

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

What is the answer to z^2-2z+1-i=0 ?
The answer to z^2-2z+1-i=0 is z=(-sqrt(2)+2)/2-1/(sqrt(2))i,z=(sqrt(2)+2)/2+1/(sqrt(2))i