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

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

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x^2-2ix-2=0

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

x = 1 + i, x = - 1 + i

Solution steps

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

Substitute

x = a + bi

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

x = a + bi

x = 1 + i, x = - 1 + i

(x + y i) + 4 (1 - i) = 5 - 7 i

z^{2} - (5 + 4 i) z + (3 - 5 i) = 0

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

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

- 8.4 x i - 3.15 = - x^{2}