If `z=((sqrt3)/(2)+(i)/(2))^(107)+((sqrt3)/(2)-(i)/(2))^(107),` then what is the imaginary pary of z equal to?

Options

A. 0
B. `1/2`
C. `(sqrt3)/(2)`
D. 1

Explanation

To solve the problem step by step, we start with the expression for \( z \):

\[ z = \left( \frac{\sqrt{3}}{2} + \frac{i}{2} \right)^{107} + \left( \frac{\sqrt{3}}{2} - \frac{i}{2} \right)^{107} \]

### Step 1: Identify the complex numbers
The complex numbers can be expressed in polar form. We note that:

\[ \frac{\sqrt{3}}{2} + \frac{i}{2} = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) \]

This means:

\[ \frac{\sqrt{3}}{2} + \frac{i}{2} = e^{i \frac{\pi}{6}} \]

Similarly, for the second term:

\[ \frac{\sqrt{3}}{2} - \frac{i}{2} = \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) = e^{-i \frac{\pi}{6}} \]

### Step 2: Rewrite \( z \) using polar form
Now we can rewrite \( z \):

\[ z = \left( e^{i \frac{\pi}{6}} \right)^{107} + \left( e^{-i \frac{\pi}{6}} \right)^{107} \]

### Step 3: Apply the exponent
Using the properties of exponents, we have:

\[ z = e^{i \frac{107\pi}{6}} + e^{-i \frac{107\pi}{6}} \]

### Step 4: Use Euler's formula
Using Euler's formula \( e^{ix} + e^{-ix} = 2 \cos(x) \):

\[ z = 2 \cos\left(\frac{107\pi}{6}\right) \]

### Step 5: Simplify the angle
Now simplify \( \frac{107\pi}{6} \):

\[ \frac{107\pi}{6} = 17\pi + \frac{5\pi}{6} = 17\pi + \frac{5\pi}{6} \equiv \frac{5\pi}{6} \quad (\text{mod } 2\pi) \]

### Step 6: Calculate the cosine
Now we can find \( \cos\left(\frac{5\pi}{6}\right) \):

\[ \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} \]

### Step 7: Substitute back into \( z \)
Substituting back, we find:

\[ z = 2 \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3} \]

### Step 8: Determine the imaginary part
Since \( z = -\sqrt{3} \) has no imaginary component, the imaginary part of \( z \) is:

\[ \text{Imaginary part of } z = 0 \]

### Final Answer
The imaginary part of \( z \) is \( 0 \).

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