If `(3+i)(z+bar(z))-(2+i)(z-bar(z))+14i=0`, where `i=sqrt(-1)`, then z `bar(z)` is equal to

Options

A. (a)10
B. (b)8
C. (c)-9
D. (d)-10

Answer

To solve the equation \((3+i)(z+\bar{z})-(2+i)(z-\bar{z})+14i=0\), where \(i=\sqrt{-1}\), we will follow these steps:

### Step 1: Rewrite \(z\) and \(\bar{z}\)
Let \(z = x + iy\), where \(x\) and \(y\) are real numbers. Then, \(\bar{z} = x - iy\).

### Step 2: Substitute \(z\) and \(\bar{z}\) into the equation
We have:
\[ z + \bar{z} = (x + iy) + (x - iy) = 2x \]
\[ z - \bar{z} = (x + iy) - (x - iy) = 2iy \]

### Step 3: Substitute into the equation
Substituting these into the original equation gives:
\[ (3+i)(2x) - (2+i)(2iy) + 14i = 0 \]

### Step 4: Expand the equation
Expanding this, we get:
\[ (6x + 2ix) - (4y + 2iy) + 14i = 0 \]
This simplifies to:
\[ 6x + 2ix - 4y - 2iy + 14i = 0 \]

### Step 5: Group real and imaginary parts
Grouping real and imaginary parts, we have:
\[ (6x - 4y) + (2x - 2y + 14)i = 0 \]
This gives us two equations:
1. \(6x - 4y = 0\)
2. \(2x - 2y + 14 = 0\)

### Step 6: Solve the first equation
From the first equation \(6x - 4y = 0\), we can express \(y\) in terms of \(x\):
\[ 4y = 6x \implies y = \frac{3}{2}x \]

### Step 7: Substitute \(y\) into the second equation
Substituting \(y = \frac{3}{2}x\) into the second equation:
\[ 2x - 2\left(\frac{3}{2}x\right) + 14 = 0 \]
This simplifies to:
\[ 2x - 3x + 14 = 0 \implies -x + 14 = 0 \implies x = 14 \]

### Step 8: Find \(y\)
Now substituting \(x = 14\) back into \(y = \frac{3}{2}x\):
\[ y = \frac{3}{2}(14) = 21 \]

### Step 9: Calculate \(z\) and \(\bar{z}\)
Now we have \(z = x + iy = 14 + 21i\) and \(\bar{z} = x - iy = 14 - 21i\).

### Step 10: Calculate \(z \bar{z}\)
Now we calculate \(z \bar{z}\):
\[ z \bar{z} = (14 + 21i)(14 - 21i) = 14^2 - (21i)^2 = 196 - 441(-1) = 196 + 441 = 637 \]

### Final Answer
Thus, \(z \bar{z} = 637\).

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