What is the answer to x^2-4y+(2y-x)i=2-i ?

The answer to x^2-4y+(2y-x)i=2-i is

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

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

(x = 2, x = 0, y = \frac{1}{2} y = - \frac{1}{2})

Solution steps

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

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

[x^{2} - 4 y = 2 2 y - x = - 1]

[x^{2} - 4 y = 2 2 y - x = - 1] : (x = 2, x = 0, y = \frac{1}{2} y = - \frac{1}{2})

(x = 2, x = 0, y = \frac{1}{2} y = - \frac{1}{2})

x^{2} - i = 0

(2 + x i) + (y + 3 i) = 7 + 4 i

(a + bi)^{2} = 3 + 4 i

∣ z - i ∣ = 1

x^{2} = 4 i