If `z=(sqrt(3)/2+i/2)^5+(sqrt(3)/2-i/2)^5` , then prove that `Im(z)=0`

Options

A. R(z) `gt 0` and `I(z) gt 0`
B. `R(z) lt 0` and `I(z) gt 0`
C. `R(z) = -3`
D. I(z) = 0

Answer

To solve the problem, we need to evaluate \( z = \left( \frac{\sqrt{3}}{2} + \frac{i}{2} \right)^5 + \left( \frac{\sqrt{3}}{2} - \frac{i}{2} \right)^5 \) and show that the imaginary part of \( z \) is zero.

### Step-by-Step Solution:

1. **Express in Polar Form:**
The complex numbers can be expressed in polar form. We recognize that:
\[ \frac{\sqrt{3}}{2} + \frac{i}{2} = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) \]
and
\[ \frac{\sqrt{3}}{2} - \frac{i}{2} = \cos\left(\frac{\pi}{6}\right) - i \sin\left(\frac{\pi}{6}\right). \]
Therefore, we can write:
\[ z = \left( \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) \right)^5 + \left( \cos\left(\frac{\pi}{6}\right) - i \sin\left(\frac{\pi}{6}\right) \right)^5. \]

2. **Use De Moivre's Theorem:**
By De Moivre's theorem:
\[ z = \left( \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) \right)^5 = \cos\left(\frac{5\pi}{6}\right) + i \sin\left(\frac{5\pi}{6}\right) \]
and
\[ \left( \cos\left(\frac{\pi}{6}\right) - i \sin\left(\frac{\pi}{6}\right) \right)^5 = \cos\left(\frac{5\pi}{6}\right) - i \sin\left(\frac{5\pi}{6}\right). \]

3. **Combine the Results:**
Thus, we can combine these results:
\[ z = \left( \cos\left(\frac{5\pi}{6}\right) + i \sin\left(\frac{5\pi}{6}\right) \right) + \left( \cos\left(\frac{5\pi}{6}\right) - i \sin\left(\frac{5\pi}{6}\right) \right). \]
This simplifies to:
\[ z = 2 \cos\left(\frac{5\pi}{6}\right). \]

4. **Evaluate \( \cos\left(\frac{5\pi}{6}\right) \):**
We know:
\[ \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}. \]
Therefore:
\[ z = 2 \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3}. \]

5. **Identify the Real and Imaginary Parts:**
The complex number \( z \) can be written as:
\[ z = -\sqrt{3} + 0i. \]
Hence, the imaginary part of \( z \) is:
\[ \text{Im}(z) = 0. \]

### Conclusion:
Thus, we have shown that \( \text{Im}(z) = 0 \).