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

The answer to x^2+2ix-1=0 is x=-i

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x^2+2ix-1=0

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

x = - i

Solution steps

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

Substitute

x = a + bi

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

x = a + bi

x = - i

z - \frac{1}{z} = 5 + 5 i

\frac{x + i y}{1 + x + i y} = 3 + 4 i

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

3 x - 2 y i = 6 + 4 i

p (2 i + 6 j) + q (2 i - 4 j) = 6 i - 7 j