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

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

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2x^2+2xi-y^2-yi-2i=2

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

(x = 1, x = 3, y = 0 y = 4)

Solution steps

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

Rewrite

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

in standard complex form:

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

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

Rewrite

2

in standard complex form:

2 + 0 i

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

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

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

[2 x^{2} - y^{2} = 2 2 x - y - 2 = 0] : (x = 1, x = 3, y = 0 y = 4)

(x = 1, x = 3, y = 0 y = 4)

a + bi = - 12 + 7 i

z^{2} - (2 + i) z - 1 + 7 i = 0

(a + bi)^{2} = - 16 - 30 i

\frac{2 + ai}{6 + bi} = 2 + 7 i

z^{2} - (1 + 4 i) z + 3 + 11 i = 0

What is the answer to 2x^2+2xi-y^2-yi-2i=2 ?
The answer to 2x^2+2xi-y^2-yi-2i=2 is