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

The answer to x^2+(1-i)x+2-2i=0 is x=2i,x=-1-i

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

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

x = 2 i, x = - 1 - i

Solution steps

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

Substitute

x = a + bi

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

Expand

(a + bi)^{2} + (1 - i) (a + bi) + 2 - 2 i : (a^{2} - b^{2} + a + b + 2) + i (- 2 - a + b + 2 ab)

(a^{2} - b^{2} + a + b + 2) + i (- 2 - a + b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} + a + b + 2) + i (- 2 - a + b + 2 ab) = 0 + 0 i

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

[a^{2} - b^{2} + a + b + 2 = 0 - 2 - a + b + 2 ab = 0]

[a^{2} - b^{2} + a + b + 2 = 0 - 2 - a + b + 2 ab = 0] : (a = 0, a = - 1, b = 2 b = - 1)

Substitute back

x = a + bi

x = 2 i, x = - 1 - i

x, cs in y + x^{2} y = c

(x + 2 y + 3) + i (3 x - y - 1) = 0

4 z^{2} + (- 13 + 18 i) z - 5 - 10 i = 0

x^{2} + 4 - 3 i = 0

- 3 ai - (- 1 - i) b = 3 a - 2

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