What is the answer to x(1-x)=y^2-iy(2x+1) ?

The answer to x(1-x)=y^2-iy(2x+1) is

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x(1-x)=y^2-iy(2x+1)

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

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

Solution steps

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

Rewrite

x (1 - x)

in standard complex form:

x (1 - x) + 0 i

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

Rewrite

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

in standard complex form:

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

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

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

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

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

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

(a^{2} - b^{2} - a + b) + (2 ab - b + a) i = 0

\frac{1 + i}{z} - (1 + 2 i) = i

f^{2} = \frac{1}{2} \cdot (f + i)

(x + 3) + 4 i = 2 + (y - 2) i

2 z^{2} - 7 z + 4 = i (2 z + 1)

What is the answer to x(1-x)=y^2-iy(2x+1) ?
The answer to x(1-x)=y^2-iy(2x+1) is