What is the answer to x^2+2ixy-y^2=0+2i ?

The answer to x^2+2ixy-y^2=0+2i is

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x^2+2ixy-y^2=0+2i

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

(x = 1, x = - 1, y = 1 y = - 1)

Solution steps

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

Rewrite

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

in standard complex form:

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

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

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

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

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

(x = 1, x = - 1, y = 1 y = - 1)

x - i y = \frac{7 - i ^{9}}{1 - i}

i (x + i y) = 2

i (x) = \frac{x ^{2}}{4} + 3450 x

4 x + 4 i y - 23 = 5 i x + 5 i^{2} y + 2 i

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