+38 Find the complex number z that satisfies:
(1+i)z-2z*=-11+25i
I beleive the i's in this problem are not variables and are the complex number representing the square root of negative one. The "z*" represents the variable "z" conjugated.
I am struggling to understand this so any help would be appreciated.
Thanks!
+1877 We are asked to find the complex number \( z \) that satisfies the equation:
\[
(1+i)z - 2z^* = -11 + 25i
\]
where \( z^* \) denotes the conjugate of \( z \). Let \( z = x + yi \), where \( x \) and \( y \) are real numbers. Then, the conjugate \( z^* \) is given by:
\[
z^* = x - yi
\]
### Step 1: Substitute \( z = x + yi \) and \( z^* = x - yi \) into the equation
Substituting these into the equation, we get:
\[
(1+i)(x + yi) - 2(x - yi) = -11 + 25i
\]
### Step 2: Expand the terms
First, expand \( (1+i)(x + yi) \):
\[
(1+i)(x + yi) = 1 \cdot x + 1 \cdot yi + i \cdot x + i \cdot yi = x + yi + xi + i^2y = (x - y) + i(x + y)
\]
Next, expand \( -2(x - yi) \):
\[
-2(x - yi) = -2x + 2yi
\]
Substituting these into the equation:
\[
(x - y + 2y)i + (x - 2x - y) = -11 + 25i
\]
### Step 3: Combine like terms
Combining the real parts and imaginary parts, we have:
\[
(x - y - 2x) + (2y + x + 2yi) = -11 + 25i
\]
This simplifies to:
\[
-x - y + 2yi + (x - 2x) + (y + y)i = -11 + 25i
\]
### Step 4: Equate real and imaginary parts
Now, equate the real and imaginary parts of the equation:
For the real part:
\[
x - 2y = -11
\]
For the imaginary part:
\[
x + 2y = 25
\]
### Step 5: Solve the system of equations
We now solve the system of equations:
\[
x - 2y = -11
\]
\[
x + 2y = 25
\]
Add these two equations:
\[
(x - 2y) + (x + 2y) = -11 + 25
\]
This simplifies to:
\[
2x = 14
\]
So:
\[
x = 7
\]
Substitute \( x = 7 \) into one of the original equations:
\[
7 - 2y = -11
\]
Solve for \( y \):
\[
-2y = -18 \quad \Rightarrow \quad y = 9
\]
### Step 6: Write the complex number \( z \)
The complex number \( z \) is:
\[
z = x + yi = 7 + 9i
\]
Thus, the solution is:
\[
\boxed{7 + 9i}
\]
