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

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

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z^2-(2+2i)z+(1+2i)=0

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

z = 1, z = 1 + 2 i

Solution steps

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

Substitute

z = x + y i

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

Expand

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

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

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

Substitute back

z = x + y i

z = 1, z = 1 + 2 i

(x + y \cdot i)^{2} = i

(x + y i) = - 8

x, S i z = \frac{x y}{x + y}

- 5 x + 3 i y = 10 - 9 i

z^{2} - (3 + 6 i) z - (9 - i) = 0

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