What is the answer to z^2-(1-2i)z+(2-16i)=0 ?

The answer to z^2-(1-2i)z+(2-16i)=0 is z=-2-4i,z=3+2i

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z^2-(1-2i)z+(2-16i)=0

z^{2} - (1 - 2 i) z + (2 - 16 i) = 0

z = - 2 - 4 i, z = 3 + 2 i

Solution steps

z^{2} - (1 - 2 i) z + (2 - 16 i) = 0

Substitute

z = x + y i

(x + y i)^{2} - (1 - 2 i) (x + y i) + 2 - 16 i = 0

Expand

(x + y i)^{2} - (1 - 2 i) (x + y i) + 2 - 16 i : (x^{2} - y^{2} - x - 2 y + 2) + i (- 16 - y + 2 x + 2 x y)

(x^{2} - y^{2} - x - 2 y + 2) + i (- 16 - y + 2 x + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - x - 2 y + 2) + i (- 16 - y + 2 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} - x - 2 y + 2 = 0 - 16 - y + 2 x + 2 x y = 0]

[x^{2} - y^{2} - x - 2 y + 2 = 0 - 16 - y + 2 x + 2 x y = 0] : (x = - 2, x = 3, y = - 4 y = 2)

Substitute back

z = x + y i

z = - 2 - 4 i, z = 3 + 2 i

(8 + 9 i) y - (3 + 6 i) x = 16 - 3 i

(5 - 3 i) y - (6 + 2 i) x = 3 - 41 i

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

a = [aij]

∣ z + 1 - i ∣ = 1

What is the answer to z^2-(1-2i)z+(2-16i)=0 ?
The answer to z^2-(1-2i)z+(2-16i)=0 is z=-2-4i,z=3+2i