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

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

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Z^2-(2+2i)Z-4+2i=0

Z^{2} - (2 + 2 i) Z - 4 + 2 i = 0

Z = 3 + i, Z = - 1 + i

Solution steps

Z^{2} - (2 + 2 i) Z - 4 + 2 i = 0

Substitute

Z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

Z = x + y i

Z = 3 + i, Z = - 1 + i

(a + bi)^{2} = 15 + 8 i

i = - 0.3 q^{2} + 30 q

x, s i x = x (x + 1) - x (x - 1)

(z - (2 + i)) (z - (2 - i)) = 0

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

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