What is the answer to z^2=-((4i)/3) ?

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

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z^2=-((4i)/3)

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

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

Solution steps

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

Substitute

z = x + y i

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

Expand

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

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

Rewrite

- \frac{4 i}{3}

in standard complex form:

0 - \frac{4}{3} i

(x^{2} - y^{2}) + 2 i x y = 0 - \frac{4}{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{4}{3}]

[x^{2} - y^{2} = 0 2 x y = - \frac{4}{3}] : x = \frac{2}{3}, x = - \frac{2}{3}, y = - \frac{2}{3} y = \frac{2}{3}

Substitute back

z = x + y i

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

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

z^{2} - (5 + 2 i) z + 1 = 0

+ 6 i z^{2} = 3 + 5 i, z^{1} = 2

(2 + 3 i) (a + bi) = 13

22 (x + y i) + (28 + 4 i) = 72 - 62 i

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