If `z= ((sqrt3)/(2) + (i)/(2))^(107) + ((sqrt3)/(2)-(i)/(2))^(107)`, then show that Im(z)=0

Answer

To solve the problem, we need to evaluate the expression:

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

### Step 1: Rewrite the terms in polar form

The first term can be expressed in polar form. We recognize that:

\[ \frac{\sqrt{3}}{2} = \cos\left(\frac{\pi}{6}\right) \quad \text{and} \quad \frac{1}{2} = \sin\left(\frac{\pi}{6}\right) \]

Thus, we can write:

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

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: Substitute into the expression for z

Now substituting back into the expression for \( z \):

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

Using the property of exponents, we can simplify this to:

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

### Step 3: 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 4: Simplify the angle

Now we need to simplify \( \frac{107\pi}{6} \). We can reduce this angle by subtracting \( 2\pi \):

\[ \frac{107\pi}{6} - 2\pi = \frac{107\pi}{6} - \frac{12\pi}{6} = \frac{95\pi}{6} \]

Continuing to reduce:

\[ \frac{95\pi}{6} - 2\pi = \frac{95\pi}{6} - \frac{12\pi}{6} = \frac{83\pi}{6} \]

Continuing this process:

\[ \frac{83\pi}{6} - 2\pi = \frac{83\pi}{6} - \frac{12\pi}{6} = \frac{71\pi}{6} \]

\[ \frac{71\pi}{6} - 2\pi = \frac{71\pi}{6} - \frac{12\pi}{6} = \frac{59\pi}{6} \]

\[ \frac{59\pi}{6} - 2\pi = \frac{59\pi}{6} - \frac{12\pi}{6} = \frac{47\pi}{6} \]

\[ \frac{47\pi}{6} - 2\pi = \frac{47\pi}{6} - \frac{12\pi}{6} = \frac{35\pi}{6} \]

\[ \frac{35\pi}{6} - 2\pi = \frac{35\pi}{6} - \frac{12\pi}{6} = \frac{23\pi}{6} \]

\[ \frac{23\pi}{6} - 2\pi = \frac{23\pi}{6} - \frac{12\pi}{6} = \frac{11\pi}{6} \]

Now we can use \( \frac{11\pi}{6} \) directly:

### Step 5: Calculate the cosine

Now we can evaluate:

\[ z = 2\cos\left(\frac{11\pi}{6}\right) \]

Since \( \cos\left(\frac{11\pi}{6}\right) = \cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \):

\[ z = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \]

### Step 6: Determine the imaginary part

Since \( z = \sqrt{3} + 0i \), we see that the imaginary part of \( z \) is:

\[ \text{Im}(z) = 0 \]

Thus, we have shown that:

\[ \text{Im}(z) = 0 \]