What is the answer to yin^0=3y^2+x-1 ?

The answer to yin^0=3y^2+x-1 is

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yin^0=3y^2+x-1

y i n^{0} = 3 y^{2} + x - 1

(y = 0, x = 1)

Solution steps

y i n^{0} = 3 y^{2} + x - 1

Rewrite

y i n^{0}

in standard complex form:

0 + y i

0 + y i = 3 y^{2} + x - 1

Rewrite

3 y^{2} + x - 1

in standard complex form:

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

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

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

[0 = 3 y^{2} + x - 1 y = 0]

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

(y = 0, x = 1)

\frac{s i}{P ( x )} = (x - 1)^{2} + x - 1

(z - \frac{3}{5})^{2} + (5 i)^{2} = 0

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

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

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