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

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

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

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

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

Solution steps

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

Substitute

z = x + y i

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

Expand

(x + y i)^{2} - (1 + 4 i) (x + y 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 + 7 y i - 4 = 5 + 21 i + 2

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

A = P (1 + \frac{i}{100})^{2}, i

z^{2} = 1 + i + \frac{58}{9 ( 3 - 7 i )}

(x + y i)^{2} = - 8 - 6 i

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