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

The answer to z^2-z-1-i=0 is z=-0.69389…-0.41879…i,z=1.69389…+0.41879…i

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z^2-z-1-i=0

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

z = - 0.69389 \dots - 0.41879 \dots i, z = 1.69389 \dots + 0.41879 \dots 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 = - 0.69389 \dots, x = 1.69389 \dots, y = - 0.41879 \dots y = 0.41879 \dots)

Substitute back

z = x + y i

z = - 0.69389 \dots - 0.41879 \dots i, z = 1.69389 \dots + 0.41879 \dots i

x, (i x - i y)^{2} = 2

(x + i y)^{2} + 2 (x - i y) = - 1 + 6 i

z^{2} = - 2 - 23 i

\frac{i}{z} = - 6

z - \frac{1}{z} = 4 + 3 i

What is the answer to z^2-z-1-i=0 ?
The answer to z^2-z-1-i=0 is z=-0.69389…-0.41879…i,z=1.69389…+0.41879…i