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

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

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

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

x = 2 + i, x = i

Solution steps

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

Substitute

x = a + bi

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

x = a + bi

x = 2 + i, x = i

(x + i y) (3 + i) = 5

(1 + i) (x + 2 y) - (3 - 2 i) (x - y) = 8 + 3 i

x + y i = 2

(x + y i) (1 + 2 i) = - 1 + 8 i

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

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