What is the answer to z^2-3+4i=0 ?

The answer to z^2-3+4i=0 is z=2-i,z=-2+i

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z^2-3+4i=0

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

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

Solution steps

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

Substitute

z = x + y i

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

Expand

(x + y i)^{2} - 3 + 4 i : (x^{2} - y^{2} - 3) + i (4 + 2 x y)

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

Rewrite

0

in standard complex form:

0 + 0 i

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

[x^{2} - y^{2} - 3 = 0 4 + 2 x y = 0] : (x = 2, x = - 2, y = - 1 y = 1)

Substitute back

z = x + y i

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

x^{2} + (2 - 2 i) x + 3 - 6 i = 0

a^{2} - b^{2} - a - bi = 1

(2 x - 3 y i) (1 - 2 i) = 1 - 4 i

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

(1 + 2 i) (x + y i) - (1 + 2 i) (x - y i) = 16