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Step-by-Step Solution
Step 1: Express each term in exponential form
Recall that
$ \frac{\sqrt{3}}{2} + \frac{i}{2} = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right). $
Using Euler's formula $ e^{i\theta} = \cos \theta + i \sin \theta, $ we can write:
$ \frac{\sqrt{3}}{2} + \frac{i}{2} = e^{i\pi/6} \quad \text{and} \quad \frac{\sqrt{3}}{2} - \frac{i}{2} = e^{-\,i\pi/6}. $
Step 2: Raise each term to the fifth power
We want to compute:
$ \left( e^{i\pi/6} \right)^5 \quad \text{and} \quad \left( e^{-\,i\pi/6} \right)^5. $
Using the property $ \left(e^{i\theta}\right)^n = e^{i n \theta}, $ we get:
$ \left( e^{i\pi/6} \right)^5 = e^{i5\pi/6}, \quad \left(e^{-\,i\pi/6}\right)^5 = e^{-\,i5\pi/6}. $
Step 3: Form the sum and simplify
Now we add the two exponential terms to find
$ z = e^{i5\pi/6} + e^{-\,i5\pi/6}.$
Using Euler's formula again,
$ e^{i\theta} = \cos\theta + i\sin\theta, $
we write:
$ e^{i5\pi/6} = \cos\left(\frac{5\pi}{6}\right) + i \sin\left(\frac{5\pi}{6}\right), \quad
e^{-\,i5\pi/6} = \cos\left(-\frac{5\pi}{6}\right) + i \sin\left(-\frac{5\pi}{6}\right). $
Adding these, we have:
$ z = \cos\left(\frac{5\pi}{6}\right) + i \sin\left(\frac{5\pi}{6}\right)
+ \cos\left(-\frac{5\pi}{6}\right) + i \sin\left(-\frac{5\pi}{6}\right). $
Since $ \cos(-\theta) = \cos(\theta) $ and $ \sin(-\theta) = -\sin(\theta), $
the imaginary parts cancel:
$ \sin\left(\frac{5\pi}{6}\right) + \sin\left(-\frac{5\pi}{6}\right) = \sin\left(\frac{5\pi}{6}\right) - \sin\left(\frac{5\pi}{6}\right) = 0. $
Hence the imaginary part of $z$ is zero, and we get:
$ z = 2\,\cos\left(\frac{5\pi}{6}\right). $
Step 4: Conclude the real and imaginary parts
1. The imaginary part is $ \mathrm{I}(z) = 0. $
2. The real part is $ \mathrm{R}(z) = 2\,\cos\left(\frac{5\pi}{6}\right). $
Note that $ \cos\left(\frac{5\pi}{6}\right) = -\,\frac{\sqrt{3}}{2} $, so
$ \mathrm{R}(z) = 2 \times \left(-\,\frac{\sqrt{3}}{2}\right) = -\sqrt{3}, $
which is less than 0.
Answer
The imaginary part $ \mathrm{I}(z) $ is zero, so the correct statement is
$ \mathrm{I}(z) = 0. $
