What is the answer to z^2=-2-2sqrt(3)i ?

The answer to z^2=-2-2sqrt(3)i is z=1-sqrt(3)i,z=-1+sqrt(3)i

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z^2=-2-2sqrt(3)i

z^{2} = - 2 - 23 i

z = 1 - 3 i, z = - 1 + 3 i

Solution steps

z^{2} = - 2 - 23 i

Substitute

z = x + y i

(x + y i)^{2} = - 2 - 23 i

Expand

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

(x^{2} - y^{2}) + 2 i x y = - 2 - 23 i

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

[x^{2} - y^{2} = - 2 2 x y = - 23]

[x^{2} - y^{2} = - 2 2 x y = - 23] : (x = 1, x = - 1, y = - 3 y = 3)

Substitute back

z = x + y i

z = 1 - 3 i, z = - 1 + 3 i

\frac{i}{z} = - 6

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

∣ i + z ∣^{2} - i - 2 = z

a^{2} = i

i x^{2} + (1 - 2 i) x + 3 i - 1 = 0

What is the answer to z^2=-2-2sqrt(3)i ?
The answer to z^2=-2-2sqrt(3)i is z=1-sqrt(3)i,z=-1+sqrt(3)i