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

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

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z^2=1+sqrt(3)i

z^{2} = 1 + 3 i

z = \frac{3}{2} + \frac{1}{2} i, z = - \frac{3}{2} - \frac{1}{2} i

Solution steps

z^{2} = 1 + 3 i

Substitute

z = x + y i

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

Expand

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

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

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

[x^{2} - y^{2} = 1 2 x y = 3]

[x^{2} - y^{2} = 1 2 x y = 3] : x = \frac{3}{2}, x = - \frac{3}{2}, y = \frac{1}{2} y = - \frac{1}{2}

Substitute back

z = x + y i

z = \frac{3}{2} + \frac{1}{2} i, z = - \frac{3}{2} - \frac{1}{2} i

(x + i y) (- i) = 14 i

1 = ai + b, 2 - i = - a + b

x + y i = 1

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

(a - bi)^{2} = 13 - 12 i

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