What is the answer to x^2-4ix-1=0 ?

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

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x^2-4ix-1=0

x^{2} - 4 i x - 1 = 0

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

Solution steps

x^{2} - 4 i x - 1 = 0

Substitute

x = a + bi

(a + bi)^{2} - 4 i (a + bi) - 1 = 0

Expand

(a + bi)^{2} - 4 i (a + bi) - 1 : (a^{2} - b^{2} + 4 b - 1) + i (- 4 a + 2 ab)

(a^{2} - b^{2} + 4 b - 1) + i (- 4 a + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} + 4 b - 1) + i (- 4 a + 2 ab) = 0 + 0 i

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

[a^{2} - b^{2} + 4 b - 1 = 0 - 4 a + 2 ab = 0]

[a^{2} - b^{2} + 4 b - 1 = 0 - 4 a + 2 ab = 0] : (a = 0, a = 0, b = 2 - 3 b = 2 + 3)

Substitute back

x = a + bi

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

z^{2} + (2 i + 7) z + 6 + 3 i = 0

y, e = \frac{sin ( x )}{x e n y}

\frac{6 - 5 i}{2 + 3 i} = a + bi

x^{2} - y^{2} - 2 y x i = 4 x + 4

x^{2} - 3 x + 1 + 3 i = 0

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