if `z_(1) = 3-i and z_(2) = -3 +i, ` then find ` Re ((z_(1)z_(2))/(barz_(1)))`

Answer

To solve the problem, we need to find the real part of the expression \(\frac{z_1 z_2}{\bar{z_1}}\) where \(z_1 = 3 - i\) and \(z_2 = -3 + i\).

### Step 1: Calculate \(z_1 z_2\)
We start by multiplying \(z_1\) and \(z_2\):

\[ z_1 z_2 = (3 - i)(-3 + i) \]

Using the distributive property (FOIL method):

\[ = 3 \cdot (-3) + 3 \cdot i - i \cdot (-3) - i \cdot i \]
\[ = -9 + 3i + 3i - i^2 \]
Since \(i^2 = -1\), we can substitute:

\[ = -9 + 6i + 1 \]
\[ = -8 + 6i \]

### Step 2: Calculate \(\bar{z_1}\)
The conjugate of \(z_1\) is:

\[ \bar{z_1} = 3 + i \]

### Step 3: Form the expression \(\frac{z_1 z_2}{\bar{z_1}}\)
Now we substitute \(z_1 z_2\) and \(\bar{z_1}\) into the expression:

\[ \frac{z_1 z_2}{\bar{z_1}} = \frac{-8 + 6i}{3 + i} \]

### Step 4: Multiply numerator and denominator by the conjugate of the denominator
To simplify, we multiply the numerator and denominator by the conjugate of the denominator, which is \(3 - i\):

\[ = \frac{(-8 + 6i)(3 - i)}{(3 + i)(3 - i)} \]

Calculating the denominator:

\[ (3 + i)(3 - i) = 3^2 - i^2 = 9 - (-1) = 10 \]

Calculating the numerator:

\[ (-8 + 6i)(3 - i) = -24 + 8i + 18i - 6i^2 \]
Substituting \(i^2 = -1\):

\[ = -24 + 26i + 6 = -18 + 26i \]

### Step 5: Combine the results
Now we have:

\[ \frac{-18 + 26i}{10} = -\frac{18}{10} + \frac{26}{10}i = -\frac{9}{5} + \frac{13}{5}i \]

### Step 6: Find the real part
The real part of the expression \(\frac{z_1 z_2}{\bar{z_1}}\) is:

\[ \text{Re}\left(\frac{z_1 z_2}{\bar{z_1}}\right) = -\frac{9}{5} \]

### Final Answer
Thus, the real part is:

\[ \boxed{-\frac{9}{5}} \]