To solve for the complex number \( z \) in the equation \( (1+i)z - 2z^* = -11 + 25i \), let’s represent \( z \) as:
\[ z = x + yi \]
where \( x \) and \( y \) are real numbers. The conjugate of \( z \) is given by:
\[ z^* = x - yi \]
Now, substituting \( z \) and \( z^* \) into the equation, we get:
\[ (1+i)(x+yi) - 2(x-yi) = -11 + 25i \]
Expanding \( (1+i)(x+yi) \):
\[ (1+i)(x+yi) = x + yi + ix - y = (x - y) + (y + x)i \]
So now we can rewrite the left side of the equation:
\[ (x - y) + (y + x)i - 2(x - yi) = -11 + 25i \] Expanding \( -2(x - yi) \):
\[ -2(x - yi) = -2x + 2yi \]
Now combining all the terms on the left side:
\[ (x - y - 2x) + (y + x + 2y)i = -11 + 25i \] This simplifies to:
\[ (-y - x) + (3y + x)i = -11 + 25i \]
We can separate the real and imaginary parts:
-
Real part: \[ -y - x = -11 \]
-
Imaginary part: \[ 3y + x = 25 \]
We now have a system of equations:
\[
- \quad -x - y = -11 \quad \Rightarrow \quad x + y = 11 \quad \text{(Equation 1)} \] \[
- \quad 3y + x = 25 \quad \text{(Equation 2)} \]
From Equation (1):
\[ x = 11 - y \]
Now substitute \( x \) into Equation (2):
\[ 3y + (11 - y) = 25 \]
Simplifying:
\[ 3y + 11 - y = 25 \] \[ 2y + 11 = 25 \] \[ 2y = 14 \] \[ y = 7 \]
Now substitute \( y \) back into Equation (1) to find \( x \):
\[ x + 7 = 11 \] \[ x = 4 \]
Thus, we have:
\[ x = 4, \quad y = 7 \]
Now we can write the complex number \( z \):
\[ z = 4 + 7i \]
Therefore, the solution is:
\[ \boxed{4 + 7i} \]