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

The answer to z^2+(1-i)z+1-i=0 is z=-0.25706…+1.52908…i,z=-0.74293…-0.52908…i

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

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

z = - 0.25706 \dots + 1.52908 \dots i, z = - 0.74293 \dots - 0.52908 \dots i

Solution steps

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

Substitute

z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

[x^{2} - y^{2} + x + y + 1 = 0 - 1 - x + y + 2 x y = 0] : (x = - 0.25706 \dots, x = - 0.74293 \dots, y = 1.52908 \dots y = - 0.52908 \dots)

Substitute back

z = x + y i

z = - 0.25706 \dots + 1.52908 \dots i, z = - 0.74293 \dots - 0.52908 \dots i

x - y i = 4 + 7 i

\frac{1}{i C} = - 0.035 i

x + i y + i x + y = 2 + i

(3 x - 2 y i)^{2} = \frac{200}{4 + 3 i}

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

What is the answer to z^2+(1-i)z+1-i=0 ?
The answer to z^2+(1-i)z+1-i=0 is z=-0.25706…+1.52908…i,z=-0.74293…-0.52908…i