Find the complex number z that satisfies:

(1+i)z-2z*=-11+25i
The notation z* stands for the conjugate of z.

1 answer

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:

  1. Real part: \[ -y - x = -11 \]

  2. Imaginary part: \[ 3y + x = 25 \]

We now have a system of equations:

\[

  1. \quad -x - y = -11 \quad \Rightarrow \quad x + y = 11 \quad \text{(Equation 1)} \] \[
  2. \quad 3y + x = 25 \quad \text{(Equation 2)} \]

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:

\[ \boxed{4 + 7i} \]