What is the answer to x^2-ix-0.5=0 ?

The answer to x^2-ix-0.5=0 is x= 1/2+1/2 i,x=-1/2+1/2 i

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x^2-ix-0.5=0

x^{2} - i x - 0.5 = 0

x = \frac{1}{2} + \frac{1}{2} i, x = - \frac{1}{2} + \frac{1}{2} i

Solution steps

x^{2} - i x - 0.5 = 0

Substitute

x = a + bi

(a + bi)^{2} - i (a + bi) - 0.5 = 0

Expand

(a + bi)^{2} - i (a + bi) - 0.5 : (a^{2} - b^{2} + b - 0.5) + i (- a + 2 ab)

(a^{2} - b^{2} + b - 0.5) + i (- a + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} + b - 0.5) + i (- 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} + b - 0.5 = 0 - a + 2 ab = 0]

[a^{2} - b^{2} + b - 0.5 = 0 - a + 2 ab = 0] : (a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{1}{2} b = \frac{1}{2})

Substitute back

x = a + bi

x = \frac{1}{2} + \frac{1}{2} i, x = - \frac{1}{2} + \frac{1}{2} i

x + i y - 4 i = 3 y - 2 i x + 9

(x + y) - 16 i = 12 + (x - y) i

z = \frac{2 + ai}{1 - i}

x + y i = 3 i - 4 + 2 - 6 i + \frac{1}{2} i

x (2 - 4 i) - 5 = x - (7 + y i)

What is the answer to x^2-ix-0.5=0 ?
The answer to x^2-ix-0.5=0 is x= 1/2+1/2 i,x=-1/2+1/2 i