If (2 +4i)² −2(x +iy) =x +iy Find x and y

Using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

$$(2+4i)^2 = 2^2 + 2(2)(4i) + (4i)^2$$ $$= 4 + 16i + 16i^2$$

Since $i^2 = -1$:

$$= 4 + 16i - 16$$ $$= -12 + 16i$$

The equation is $(2 +4i)^2 -2(x +iy) =x +iy$.

$$(-12 + 16i) - 2(x + iy) = x + iy$$

For the real parts:

$$-12 - 2x = x$$

For the imaginary parts:

$$16 - 2y = y$$

From the real parts equation:

$$-12 = x + 2x$$ $$-12 = 3x$$ $$x = \frac{-12}{3}$$ $$x = -4$$

From the imaginary parts equation:

$$16 = y + 2y$$ $$16 = 3y$$ $$y = \frac{16}{3}$$

The values are $x=-4$ and $y=\frac{16}{3}$.

The final answer is $x=-4, y=\frac{16}{3}$.