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

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

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

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

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

Solution steps

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

Rewrite

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

in standard complex form:

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

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

Rewrite

i

in standard complex form:

0 + i

(x^{2} - y^{2}) + 2 y x i = 0 + 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 y x = 1]

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

∣ z ∣^{2} + z = 11 + 3 i

z^{2} + ∣ z ∣^{2} = 50 + 3 i

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

5 + bi = (a + b i^{2})

z, (2 - i) z + 8 z^{2} = 0