What is the answer to x^2-3x+3+i=0 ?

The answer to x^2-3x+3+i=0 is x=1+i,x=2-i

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x^2-3x+3+i=0

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

x = 1 + i, x = 2 - i

Solution steps

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

Substitute

x = a + bi

(a + bi)^{2} - 3 (a + bi) + 3 + i = 0

Expand

(a + bi)^{2} - 3 (a + bi) + 3 + i : (a^{2} - b^{2} - 3 a + 3) + i (1 - 3 b + 2 ab)

(a^{2} - b^{2} - 3 a + 3) + i (1 - 3 b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} - 3 a + 3) + i (1 - 3 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} - 3 a + 3 = 0 1 - 3 b + 2 ab = 0]

[a^{2} - b^{2} - 3 a + 3 = 0 1 - 3 b + 2 ab = 0] : (a = 1, a = 2, b = 1 b = - 1)

Substitute back

x = a + bi

x = 1 + i, x = 2 - i

(a + 5) + (b - 1) i = 7 - 3 i

z^{2} + 4 i z + 5 = 0

\frac{2 + 5 i}{1 - i} = x + y i

a, ∣ a + bi ∣ = 1

a, ∣ a + bi ∣ = 3