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

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

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z^2=-(2/3)i

z^{2} = - (\frac{2}{3}) i

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

Solution steps

z^{2} = - (\frac{2}{3}) i

Substitute

z = x + y i

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

Expand

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

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

Rewrite

- \frac{2}{3} i

in standard complex form:

0 - \frac{2}{3} i

(x^{2} - y^{2}) + 2 i x y = 0 - \frac{2}{3} i

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

[x^{2} - y^{2} = 0 2 x y = - \frac{2}{3}]

[x^{2} - y^{2} = 0 2 x y = - \frac{2}{3}] : x = \frac{1}{3}, x = - \frac{1}{3}, y = - \frac{3 \frac{2}{3}}{2} y = \frac{3 \frac{2}{3}}{2}

Substitute back

z = x + y i

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

x + i y + \frac{1}{x + i y} = 0

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

(a^{2} - b^{2} - a + b) + (2 ab - b + a) i = 0

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

f^{2} = \frac{1}{2} \cdot (f + i)

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