To solve for the number that is not part of the solution set for the equation \( z^5 \neq 7 \), we first need to understand what this inequality implies.
The inequality \( z^5 \neq 7 \) means that \( z^5 \) can be any value except for 7. Therefore, the solution set includes all complex numbers \( z \) except those for which \( z^5 = 7 \).
To find the specific values of \( z \) that satisfy \( z^5 = 7 \), we can take the fifth root of 7. The principal root is:
\[ z = \sqrt[5]{7} \]
However, since the equation is polynomial and has complex solutions, we can express all roots using De Moivre's Theorem. The fifth roots of 7 can be expressed as:
\[ z_k = \sqrt[5]{7} \cdot \text{cis}\left(\frac{2k\pi}{5}\right) \quad \text{for } k = 0, 1, 2, 3, 4 \]
where \( \text{cis}(\theta) = \cos(\theta) + i\sin(\theta) \).
Thus, there are five specific complex values of \( z \) (the 5th roots of 7) that are not part of the solution set for \( z^5 \neq 7\).
In conclusion, any number \( z \) such that \( z^5 = 7 \) is not part of the solution set. The numbers that are excluded from \( z^5 \neq 7 \) are the five distinct fifth roots of 7, specifically:
- \( z_0 = \sqrt[5]{7} \) (the principal root)
- \( z_1 = \sqrt[5]{7} \cdot \text{cis}\left(\frac{2\pi}{5}\right) \)
- \( z_2 = \sqrt[5]{7} \cdot \text{cis}\left(\frac{4\pi}{5}\right) \)
- \( z_3 = \sqrt[5]{7} \cdot \text{cis}\left(\frac{6\pi}{5}\right) \)
- \( z_4 = \sqrt[5]{7} \cdot \text{cis}\left(\frac{8\pi}{5}\right) \)
If you were to ask about a specific number, then \(\sqrt[5]{7}\) (approximately 1.903) is certainly one example of a number that is not in the solution set.