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

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

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z^2-(5+4i)z-(3-5i)=0

z^{2} - (5 + 4 i) z - (3 - 5 i) = 0

z = i, z = 5 + 3 i

Solution steps

z^{2} - (5 + 4 i) z - (3 - 5 i) = 0

Substitute

z = x + y i

(x + y i)^{2} - (5 + 4 i) (x + y i) - (3 - 5 i) = 0

Expand

(x + y i)^{2} - (5 + 4 i) (x + y i) - (3 - 5 i) : (x^{2} - y^{2} - 5 x + 4 y - 3) + i (5 - 4 x - 5 y + 2 x y)

(x^{2} - y^{2} - 5 x + 4 y - 3) + i (5 - 4 x - 5 y + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 5 x + 4 y - 3) + i (5 - 4 x - 5 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} - 5 x + 4 y - 3 = 0 5 - 4 x - 5 y + 2 x y = 0]

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

Substitute back

z = x + y i

z = i, z = 5 + 3 i

s i f (x) = x^{2} - x, f (x + 1) = f (- x)

(5 x + 4) + 7 i = (3 y + 1) i + 9

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

(1 + 5 i) p - 2 q = 3 + 7 i

x + y i = 3 - 2 i

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