The problem asks to solve the complex equation $2z - (1+i)\overline{z} = -1 + 5i$ for $z$.

1. Problem Description

The problem asks to solve the complex equation $2z - (1+i)\overline{z} = -1 + 5i$ for $z$.

2. Solution Steps

Let $z = x + iy$, where $x$ and $y$ are real numbers. Then the complex conjugate of $z$ is $\overline{z} = x - iy$.

Substitute $z = x + iy$ and $\overline{z} = x - iy$ into the equation:

$2(x + iy) - (1+i)(x - iy) = -1 + 5i$

$2x + 2iy - (x - iy + ix - i^2y) = -1 + 5i$

Since $i^2 = -1$, we have:

$2x + 2iy - (x - iy + ix + y) = -1 + 5i$

$2x + 2iy - x + iy - ix - y = -1 + 5i$

$(2x - x - y) + (2y + y - x)i = -1 + 5i$

$(x - y) + (3y - x)i = -1 + 5i$

Now, equate the real and imaginary parts:

$x - y = -1$

$3y - x = 5$

We now have a system of two linear equations with two variables:

$x - y = -1$ (1)

$-x + 3y = 5$ (2)

Adding equations (1) and (2), we get:

$x - y - x + 3y = -1 + 5$

$2y = 4$

$y = 2$

Substitute $y = 2$ into equation (1):

$x - 2 = -1$

$x = -1 + 2$

$x = 1$

Therefore, $z = x + iy = 1 + 2i$.

3. Final Answer

$z = 1 + 2i$