What is the answer to -i=x^2-y^2+2xyi ?

The answer to -i=x^2-y^2+2xyi is

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-i=x^2-y^2+2xyi

- i = x^{2} - y^{2} + 2 x y i

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

Solution steps

- i = x^{2} - y^{2} + 2 x y i

Rewrite

- i

in standard complex form:

0 - i

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

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

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

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

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

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

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

x + 1 + i (y + 1) = 2

z^{2} = 2 + i

a - bi + (1 - i) = 3 + 2 i