If `(7+i)(z + bar(z))-(4+i)(z-bar(z)) + 116i = 0` then `z bar(z)` is equal to ______________.

Answer

To solve the equation \((7+i)(z + \bar{z})-(4+i)(z-\bar{z}) + 116i = 0\) and find the value of \(z \bar{z}\), we can follow these steps:

### Step 1: Substitute \(z\) with \(x + iy\)
Let \(z = x + iy\), where \(x\) is the real part and \(y\) is the imaginary part. The conjugate of \(z\) is \(\bar{z} = x - iy\).

### Step 2: Rewrite the equation
Substituting \(z\) and \(\bar{z}\) into the equation gives:
\[ (7+i)((x + iy) + (x - iy)) - (4+i)((x + iy) - (x - iy)) + 116i = 0 \]
This simplifies to:
\[ (7+i)(2x) - (4+i)(2iy) + 116i = 0 \]

### Step 3: Expand the equation
Expanding both terms:
\[ (14x + 2xi) - (8y + 2yi) + 116i = 0 \]
This can be grouped into real and imaginary parts:
\[ 14x - 8y + (2x - 2y + 116)i = 0 \]

### Step 4: Set real and imaginary parts to zero
From the equation, we can set the real part and the imaginary part to zero:
1. \(14x - 8y = 0\) (Real part)
2. \(2x - 2y + 116 = 0\) (Imaginary part)

### Step 5: Solve the system of equations
From the first equation, we can express \(y\) in terms of \(x\):
\[ y = \frac{14}{8}x = \frac{7}{4}x \]
Substituting \(y\) into the second equation:
\[ 2x - 2\left(\frac{7}{4}x\right) + 116 = 0 \]
This simplifies to:
\[ 2x - \frac{14}{4}x + 116 = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ 8x - 14x + 464 = 0 \]
This simplifies to:
\[ -6x + 464 = 0 \implies 6x = 464 \implies x = \frac{464}{6} = \frac{232}{3} \]

### Step 6: Find \(y\)
Now substituting \(x\) back to find \(y\):
\[ y = \frac{7}{4} \cdot \frac{232}{3} = \frac{1624}{12} = \frac{406}{3} \]

### Step 7: Calculate \(z \bar{z}\)
Now we calculate \(z \bar{z}\):
\[ z \bar{z} = (x + iy)(x - iy) = x^2 + y^2 \]
Calculating \(x^2\) and \(y^2\):
\[ x^2 = \left(\frac{232}{3}\right)^2 = \frac{53824}{9}, \quad y^2 = \left(\frac{406}{3}\right)^2 = \frac{164836}{9} \]
Thus,
\[ z \bar{z} = \frac{53824 + 164836}{9} = \frac{218660}{9} \]

### Final Answer
The value of \(z \bar{z}\) is \(\frac{218660}{9}\).