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

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

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

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

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

Solution steps

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

Substitute

z = x + y i

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

Expand

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

Rewrite

\frac{2 i}{3}

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{1}{3} y = - \frac{1}{3})

Substitute back

z = x + y i

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

z^{2} + 3 i z + 4 = 0

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

5 x + 7 y i = - x - 7 i

z^{2} - (1 - 2 i) z + (2 + 14 i) = 0

3 b - 2 ai = 4 - 3 i

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