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

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

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x^2+2ix+1=0

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

x = (- 1 - 2) i, x = (2 - 1) 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, a = 0, b = - 1 - 2 b = 2 - 1)

Substitute back

x = a + bi

x = (- 1 - 2) i, x = (2 - 1) i

2 z^{2} + (- 1 + i) z + 3 + i = 0

a + b - 2 abi = 5 - 12 i

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

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

\frac{a + 7 i}{3 - 2 i} = 1 + bi

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