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

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

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

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

z = 2 - i, z = - i

Solution steps

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

Substitute

z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

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

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

Substitute back

z = x + y i

z = 2 - i, z = - i

x^{2} + x + i = 0

- 12 x + y i = 60 - 13 i

z^{2} - (6 - 2 i) z + (11 - 10 i) = 0

i = 1.4 x 1 0^{- 4} + 4.363 x 1 0^{- 4} i

i w^{2} + 8 w - 7 i = 0

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