What is the answer to z^2=4(-1-isqrt(3)) ?

The answer to z^2=4(-1-isqrt(3)) is z=sqrt(2)-sqrt(6)i,z=-sqrt(2)+sqrt(6)i

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z^2=4(-1-isqrt(3))

z^{2} = 4 (- 1 - i 3)

z = 2 - 6 i, z = - 2 + 6 i

Solution steps

z^{2} = 4 (- 1 - i 3)

Substitute

z = x + y i

(x + y i)^{2} = 4 (- 1 - i 3)

Expand

(x^{2} - y^{2}) + 2 i x y = - 4 - 43 i

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

[x^{2} - y^{2} = - 4 2 x y = - 43]

[x^{2} - y^{2} = - 4 2 x y = - 43] : (x = 2, x = - 2, y = - 6 y = 6)

Substitute back

z = x + y i

z = 2 - 6 i, z = - 2 + 6 i

i (t) = 4 t^{2} - 48 t + 154

i (t) = 8 t^{2} - 4 t A

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

∣ z + i ∣ = 3 + 0 i

x^{2} - 3 (1 - i) x - 5 i = 0

What is the answer to z^2=4(-1-isqrt(3)) ?
The answer to z^2=4(-1-isqrt(3)) is z=sqrt(2)-sqrt(6)i,z=-sqrt(2)+sqrt(6)i