What is the answer to z^2+(2-12i)z+(-1-9i)=0 ?

The answer to z^2+(2-12i)z+(-1-9i)=0 is z=-0.74300…+0.16338…i,z=-1.25699…+11.83661…i

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z^2+(2-12i)z+(-1-9i)=0

z^{2} + (2 - 12 i) z + (- 1 - 9 i) = 0

z = - 0.74300 \dots + 0.16338 \dots i, z = - 1.25699 \dots + 11.83661 \dots i

Solution steps

z^{2} + (2 - 12 i) z + (- 1 - 9 i) = 0

Substitute

z = x + y i

(x + y i)^{2} + (2 - 12 i) (x + y i) - 1 - 9 i = 0

Expand

(x + y i)^{2} + (2 - 12 i) (x + y i) - 1 - 9 i : (x^{2} - y^{2} + 2 x + 12 y - 1) + i (- 9 - 12 x + 2 y + 2 x y)

(x^{2} - y^{2} + 2 x + 12 y - 1) + i (- 9 - 12 x + 2 y + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} + 2 x + 12 y - 1) + i (- 9 - 12 x + 2 y + 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} + 2 x + 12 y - 1 = 0 - 9 - 12 x + 2 y + 2 x y = 0]

[x^{2} - y^{2} + 2 x + 12 y - 1 = 0 - 9 - 12 x + 2 y + 2 x y = 0] : (x = - 0.74300 \dots, x = - 1.25699 \dots, y = 0.16338 \dots y = 11.83661 \dots)

Substitute back

z = x + y i

z = - 0.74300 \dots + 0.16338 \dots i, z = - 1.25699 \dots + 11.83661 \dots i

x, s i g = \frac{2}{x - 3}

z^{2} (1 + i) = 2 i

z = z + \frac{i}{z}

s i (x + 3)^{2} = (x - 1)^{2} + 18

x, f s i f (x) = x^{2} - 3 x + 2

What is the answer to z^2+(2-12i)z+(-1-9i)=0 ?
The answer to z^2+(2-12i)z+(-1-9i)=0 is z=-0.74300…+0.16338…i,z=-1.25699…+11.83661…i