What is the answer to z^2+4i=0 ?

The answer to z^2+4i=0 is z=sqrt(2)-sqrt(2)i,z=-sqrt(2)+sqrt(2)i

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z^2+4i=0

z^{2} + 4 i = 0

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

Solution steps

z^{2} + 4 i = 0

Substitute

z = x + y i

(x + y i)^{2} + 4 i = 0

Expand

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

(x^{2} - y^{2}) + i (4 + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2}) + i (4 + 2 x y) = 0 + 0 i

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

[x^{2} - y^{2} = 0 4 + 2 x y = 0]

[x^{2} - y^{2} = 0 4 + 2 x y = 0] : (x = 2, x = - 2, y = - 2 y = 2)

Substitute back

z = x + y i

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

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

\frac{a + 2 i}{3 + bi} = 3

(a + bi) (x + y i) = 2 + 2 i

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

(a + ib)^{2} = - 5 + 12 i

What is the answer to z^2+4i=0 ?
The answer to z^2+4i=0 is z=sqrt(2)-sqrt(2)i,z=-sqrt(2)+sqrt(2)i