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

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

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x^2-3ix+(-3+i)=0

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

x = 1 + i, x = - 1 + 2 i

Solution steps

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

Substitute

x = a + bi

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

x = a + bi

x = 1 + i, x = - 1 + 2 i

z^{2} - (3 + 6 i) z - (4 + 6 i) = 0

25 - 49 i = x^{2} + y^{2} i

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

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

x^{2} + (2 - 2 i) x + 3 - 6 i = 0

What is the answer to x^2-3ix+(-3+i)=0 ?
The answer to x^2-3ix+(-3+i)=0 is x=1+i,x=-1+2i