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

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

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z^2-2iz-1=0

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

z = i

Solution steps

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

Substitute

z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

z = x + y i

z = i

1 - 27 k_{1} i - 8 k_{2} = 0

z^{2} - (2 + i) \cdot z + (- 1 + 7 i) = 0

z, z^{2} = 1 + i

x i (x + 3 y) = (1 + y i) (3 x + 4 i)

x^{2} - 3 - 3 i = 0