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

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

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z^2+z+1-i=0

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

z = i, z = - 1 - i

Solution steps

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

Substitute

z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

z = x + y i

z = i, z = - 1 - i

x + 2 + 3 y i = - 4 x + (y + 4) i

(2 + i) x - y = 0

i z^{2} + 2 z = z + i

(x^{2} - 9) + (2 y - 5) i = 7 i

x + 3 i + 3 = 5 + y i