What is the answer to x^2-3(1-i)x-5i=0 ?

The answer to x^2-3(1-i)x-5i=0 is x=1-2i,x=2-i

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x^2-3(1-i)x-5i=0

x^{2} - 3 (1 - i) x - 5 i = 0

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

Solution steps

x^{2} - 3 (1 - i) x - 5 i = 0

Substitute

x = a + bi

(a + bi)^{2} - 3 (1 - i) (a + bi) - 5 i = 0

Expand

(a + bi)^{2} - 3 (1 - i) (a + bi) - 5 i : (a^{2} - b^{2} - 3 a - 3 b) + i (- 5 + 3 a - 3 b + 2 ab)

(a^{2} - b^{2} - 3 a - 3 b) + i (- 5 + 3 a - 3 b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} - 3 a - 3 b) + i (- 5 + 3 a - 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 b = 0 - 5 + 3 a - 3 b + 2 ab = 0]

[a^{2} - b^{2} - 3 a - 3 b = 0 - 5 + 3 a - 3 b + 2 ab = 0] : (a = 1, a = 2, b = - 2 b = - 1)

Substitute back

x = a + bi

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

\frac{( x - 1 )}{2 - x} = i

x y + (x - y) i = 4

∣ z - 1 - i ∣ = 1

x^{2} + 2 (1 + i) x - 5 (1 + 2 i) = 0

i (x) = \frac{640000}{4 - 4}

What is the answer to x^2-3(1-i)x-5i=0 ?
The answer to x^2-3(1-i)x-5i=0 is x=1-2i,x=2-i