What is the answer to z^2-(3-4i)*z+(2-8i)=0 ?

The answer to z^2-(3-4i)*z+(2-8i)=0 is z=1-4i,z=2

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z^2-(3-4i)*z+(2-8i)=0

z^{2} - (3 - 4 i) \cdot z + (2 - 8 i) = 0

z = 1 - 4 i, z = 2

Solution steps

z^{2} - (3 - 4 i) z + (2 - 8 i) = 0

Substitute

z = x + y i

(x + y i)^{2} - (3 - 4 i) (x + y i) + 2 - 8 i = 0

Expand

(x + y i)^{2} - (3 - 4 i) (x + y i) + 2 - 8 i : (x^{2} - y^{2} - 3 x - 4 y + 2) + i (- 8 - 3 y + 4 x + 2 x y)

(x^{2} - y^{2} - 3 x - 4 y + 2) + i (- 8 - 3 y + 4 x + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 3 x - 4 y + 2) + i (- 8 - 3 y + 4 x + 2 x y) = 0 + 0 i

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

[x^{2} - y^{2} - 3 x - 4 y + 2 = 0 - 8 - 3 y + 4 x + 2 x y = 0]

[x^{2} - y^{2} - 3 x - 4 y + 2 = 0 - 8 - 3 y + 4 x + 2 x y = 0] : (x = 1, x = 2, y = - 4 y = 0)

Substitute back

z = x + y i

z = 1 - 4 i, z = 2

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

∣ z - 3 + 4 i ∣ = 5

(3 - 2 i) (x + i) = 17 + i y

z^{2} + z \cdot i + 2 = 0

(a + 4 i) + (7 + bi) = 5 + 7 i

What is the answer to z^2-(3-4i)*z+(2-8i)=0 ?
The answer to z^2-(3-4i)*z+(2-8i)=0 is z=1-4i,z=2