What is the answer to z^2-(1+4i)z-(6-12i)=0 ?

The answer to z^2-(1+4i)z-(6-12i)=0 is z=-2+4i,z=3

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z^2-(1+4i)z-(6-12i)=0

z^{2} - (1 + 4 i) z - (6 - 12 i) = 0

z = - 2 + 4 i, z = 3

Solution steps

z^{2} - (1 + 4 i) z - (6 - 12 i) = 0

Substitute

z = x + y i

(x + y i)^{2} - (1 + 4 i) (x + y i) - (6 - 12 i) = 0

Expand

(x + y i)^{2} - (1 + 4 i) (x + y i) - (6 - 12 i) : (x^{2} - y^{2} - x + 4 y - 6) + i (12 - y - 4 x + 2 x y)

(x^{2} - y^{2} - x + 4 y - 6) + i (12 - y - 4 x + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - x + 4 y - 6) + i (12 - 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} - x + 4 y - 6 = 0 12 - y - 4 x + 2 x y = 0]

[x^{2} - y^{2} - x + 4 y - 6 = 0 12 - y - 4 x + 2 x y = 0] : (x = - 2, x = 3, y = 4 y = 0)

Substitute back

z = x + y i

z = - 2 + 4 i, z = 3

(2 + i) \cdot a + (i - 3) \cdot b = 4

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

\frac{x + i y}{1 + ( x - i y )} = 3 + 4 i

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

z - \frac{1}{z} = 3 + 4 i

What is the answer to z^2-(1+4i)z-(6-12i)=0 ?
The answer to z^2-(1+4i)z-(6-12i)=0 is z=-2+4i,z=3