What is the answer to z^2+4iz-1=0 ?

The answer to z^2+4iz-1=0 is z=(-2-sqrt(3))i,z=(-2+sqrt(3))i

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z^2+4iz-1=0

z^{2} + 4 i z - 1 = 0

z = (- 2 - 3) i, z = (- 2 + 3) i

Solution steps

z^{2} + 4 i z - 1 = 0

Substitute

z = x + y i

(x + y i)^{2} + 4 i (x + y i) - 1 = 0

Expand

(x + y i)^{2} + 4 i (x + y i) - 1 : (x^{2} - y^{2} - 4 y - 1) + i (4 x + 2 x y)

(x^{2} - y^{2} - 4 y - 1) + i (4 x + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 4 y - 1) + i (4 x + 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} - 4 y - 1 = 0 4 x + 2 x y = 0]

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

Substitute back

z = x + y i

z = (- 2 - 3) i, z = (- 2 + 3) i

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

z^{2} = - 15 + 8 i

\frac{( - 6 i + n - 3 n + 5 )}{4 + n ^{2}} = 0

(a + 2 i) (1 + 2 i) = 4 - bi

z^{2} - \frac{2 i + 1}{2 - i} = 0

What is the answer to z^2+4iz-1=0 ?
The answer to z^2+4iz-1=0 is z=(-2-sqrt(3))i,z=(-2+sqrt(3))i