Find number of values of complex numbers `omega` satisfying the system of equaltion `z^(3)=-(bar(omega))^(7)` and `z^(5).omega^(11)=1`

Explanation

To solve the given system of equations for the complex number \(\omega\), we have:

1. \( z^3 = -(\bar{\omega})^7 \) (Equation 1)
2. \( z^5 \cdot \omega^{11} = 1 \) (Equation 2)

### Step 1: Express \( z \) in terms of \( \omega \)

From Equation 2, we can express \( z \) as follows:

\[ z^5 = \frac{1}{\omega^{11}} \]

Taking the fifth root of both sides gives:

\[ z = \left(\frac{1}{\omega^{11}}\right)^{1/5} = \frac{1}{\omega^{11/5}} \cdot e^{\frac{2\pi i k}{5}} \quad \text{for } k = 0, 1, 2, 3, 4 \]

This means \( z \) can take on 5 different values based on the fifth roots of unity.

### Step 2: Substitute \( z \) into Equation 1

Now we substitute \( z \) into Equation 1:

\[ \left(\frac{1}{\omega^{11/5}} \cdot e^{\frac{2\pi i k}{5}}\right)^3 = -(\bar{\omega})^7 \]

This simplifies to:

\[ \frac{1}{\omega^{33/5}} \cdot e^{\frac{6\pi i k}{5}} = -(\bar{\omega})^7 \]

### Step 3: Rewrite \(-(\bar{\omega})^7\)

Recall that \(\bar{\omega} = \frac{1}{\omega}\) when \(\omega\) is on the unit circle. Thus:

\[ -(\bar{\omega})^7 = -\left(\frac{1}{\omega}\right)^7 = -\frac{1}{\omega^7} \]

### Step 4: Set up the equation

Now we have:

\[ \frac{1}{\omega^{33/5}} \cdot e^{\frac{6\pi i k}{5}} = -\frac{1}{\omega^7} \]

Multiplying both sides by \(-\omega^{33/5}\):

\[ e^{\frac{6\pi i k}{5}} = -\omega^{7 - \frac{33}{5}} = -\omega^{\frac{-8}{5}} \]

### Step 5: Solve for \(\omega\)

This implies:

\[ \omega^{\frac{-8}{5}} = -e^{\frac{6\pi i k}{5}} \]

Taking the reciprocal gives:

\[ \omega^{\frac{8}{5}} = -e^{\frac{6\pi i k}{5}} \]

### Step 6: Find the number of solutions

The left side, \(\omega^{\frac{8}{5}}\), has 5 distinct values for each \(k\) (since \(k\) can take values 0, 1, 2, 3, 4). Therefore, for each \(k\), we can find a corresponding \(\omega\).

### Conclusion

Thus, the total number of distinct values of \(\omega\) satisfying the system of equations is 5.