Question

Find the complex number z that satisfies:
(1+i)z-2z*=-11+25i
The notation z* stands for the conjugate of z.

Answer 1

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:

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