What is the answer to z^2-z(1+4i)+3+11i=0 ?

The answer to z^2-z(1+4i)+3+11i=0 is z=-1+5i,z=2-i

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z^2-z(1+4i)+3+11i=0

z^{2} - z (1 + 4 i) + 3 + 11 i = 0

z = - 1 + 5 i, z = 2 - i

Solution steps

z^{2} - z (1 + 4 i) + 3 + 11 i = 0

Substitute

z = x + y i

(x + y i)^{2} - (x + y i) (1 + 4 i) + 3 + 11 i = 0

Expand

(x + y i)^{2} - (x + y i) (1 + 4 i) + 3 + 11 i : (x^{2} - y^{2} - x + 4 y + 3) + i (11 - y - 4 x + 2 x y)

(x^{2} - y^{2} - x + 4 y + 3) + i (11 - y - 4 x + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - x + 4 y + 3) + i (11 - 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 + 3 = 0 11 - y - 4 x + 2 x y = 0]

[x^{2} - y^{2} - x + 4 y + 3 = 0 11 - y - 4 x + 2 x y = 0] : (x = - 1, x = 2, y = 5 y = - 1)

Substitute back

z = x + y i

z = - 1 + 5 i, z = 2 - i

x^{2} - 6 i^{2} = 0

2 x^{2} + 9 x - i = 0

z^{2} = (1 + i 3)^{2}

\frac{1 + ai}{6 + bi} = 1 + 5 i

0 = \frac{1 + λi}{1 - λi}

What is the answer to z^2-z(1+4i)+3+11i=0 ?
The answer to z^2-z(1+4i)+3+11i=0 is z=-1+5i,z=2-i