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

The answer to z^2+2z-i=0 is z=0.09868…+0.45508…i,z=-2.09868…-0.45508…i

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z^2+2z-i=0

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

z = 0.09868 \dots + 0.45508 \dots i, z = - 2.09868 \dots - 0.45508 \dots i

Solution steps

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

Substitute

z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

[x^{2} - y^{2} + 2 x = 0 - 1 + 2 y + 2 x y = 0] : (x = 0.09868 \dots, x = - 2.09868 \dots, y = 0.45508 \dots y = - 0.45508 \dots)

Substitute back

z = x + y i

z = 0.09868 \dots + 0.45508 \dots i, z = - 2.09868 \dots - 0.45508 \dots i

\frac{a + 7 i}{3 - 2 i} = 1 + bi

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

a - bi - 1 = \frac{15 + 8 i}{4 ( a + bi )}

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

\frac{2 i z}{1 + i} = \frac{i}{- z}

What is the answer to z^2+2z-i=0 ?
The answer to z^2+2z-i=0 is z=0.09868…+0.45508…i,z=-2.09868…-0.45508…i