Given the equation (2+4i)2 − 2(x+iy) = x+iy, find the values of x and y from the options below.

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 $i^2 = -1$:

$$= 4 + 16i + 16(-1)$$ $$= 4 + 16i - 16$$ $$= -12 + 16i$$

The given equation is $(2+4i)^2 - 2(x+iy) = x+iy$.
Substitute the expanded term:

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

For the real parts:

$$-12 - 2x = x \quad (1)$$

For the imaginary parts:

$$16 - 2y = y \quad (2)$$

From equation (1):

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

From equation (2):

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

The values are $x=-4$ and $y=\frac{16}{3}$.
Looking at the options:
A. $-4, \frac{16}{3}$
B. $-4, 5$
C. $4, -3$
D. $2, 16$

The calculated values match option A.

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