What is the answer to z^2-(7-i)z+(14-5i)=0 ?

The answer to z^2-(7-i)z+(14-5i)=0 is z=3-2i,z=4+i

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z^2-(7-i)z+(14-5i)=0

z^{2} - (7 - i) z + (14 - 5 i) = 0

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

Solution steps

z^{2} - (7 - i) z + (14 - 5 i) = 0

Substitute

z = x + y i

(x + y i)^{2} - (7 - i) (x + y i) + 14 - 5 i = 0

Expand

(x + y i)^{2} - (7 - i) (x + y i) + 14 - 5 i : (x^{2} - y^{2} - 7 x - y + 14) + i (- 5 + x - 7 y + 2 x y)

(x^{2} - y^{2} - 7 x - y + 14) + i (- 5 + x - 7 y + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 7 x - y + 14) + i (- 5 + x - 7 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} - 7 x - y + 14 = 0 - 5 + x - 7 y + 2 x y = 0]

[x^{2} - y^{2} - 7 x - y + 14 = 0 - 5 + x - 7 y + 2 x y = 0] : (x = 3, x = 4, y = - 2 y = 1)

Substitute back

z = x + y i

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

2 x + 2 y = 2 x + 2 y i + 5

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

- z^{2} = 3 + i

3 a - 2 bi = 5 + 3 i

22 + i 22 = x^{2}

What is the answer to z^2-(7-i)z+(14-5i)=0 ?
The answer to z^2-(7-i)z+(14-5i)=0 is z=3-2i,z=4+i