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

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

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

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

x = 2 - i, x = - i

Solution steps

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

Substitute

x = a + bi

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

[- 2 ab - 2 a + 2 b + 2 = 0 - 1 + a^{2} - b^{2} - 2 a - 2 b = 0]

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

Substitute back

x = a + bi

x = 2 - i, x = - i

(x + y i) - (- 8 + 11 i) = 5 + 9 i

z^{2} = 33 - 56 i

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

\frac{5 - 2 i}{z} = 2 + i

(x - y i)^{2} = 4 (x + y i)

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