If `Z=(7+i)/(3+4i)` , then find `Z^14`.

Options

A. `2^7`
B. `(-2)^7`
C. `(2^7) i`
D. `(-2^7) i`

Answer

To find \( Z^{14} \) for \( Z = \frac{7+i}{3+4i} \), we will follow these steps:

### Step 1: Rationalize the denominator
We start by multiplying the numerator and denominator by the conjugate of the denominator:

\[ Z = \frac{7+i}{3+4i} \cdot \frac{3-4i}{3-4i} = \frac{(7+i)(3-4i)}{(3+4i)(3-4i)} \]

### Step 2: Calculate the denominator
Using the formula \( (a+b)(a-b) = a^2 - b^2 \):

\[ (3+4i)(3-4i) = 3^2 - (4i)^2 = 9 - 16(-1) = 9 + 16 = 25 \]

### Step 3: Calculate the numerator
Now we calculate the numerator:

\[ (7+i)(3-4i) = 21 - 28i + 3i - 4i^2 = 21 - 25i + 4 = 25 - 25i \]

### Step 4: Combine the results
Now we can write \( Z \):

\[ Z = \frac{25 - 25i}{25} = 1 - i \]

### Step 5: Convert to polar form
Next, we convert \( Z \) to polar form. The modulus \( r \) and argument \( \theta \) are calculated as follows:

\[ r = |Z| = \sqrt{(1)^2 + (-1)^2} = \sqrt{2} \]

\[ \theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \]

Thus, we can express \( Z \) in polar form:

\[ Z = \sqrt{2} \left( \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right) \right) \]

### Step 6: Find \( Z^{14} \)
Using De Moivre's theorem:

\[ Z^{14} = \left( \sqrt{2} \right)^{14} \left( \cos\left(14 \cdot -\frac{\pi}{4}\right) + i \sin\left(14 \cdot -\frac{\pi}{4}\right) \right) \]

Calculating \( \left( \sqrt{2} \right)^{14} \):

\[ \left( \sqrt{2} \right)^{14} = (2^{1/2})^{14} = 2^{7} = 128 \]

Now calculate the angle:

\[ 14 \cdot -\frac{\pi}{4} = -\frac{14\pi}{4} = -\frac{7\pi}{2} \]

### Step 7: Simplify the angle
To simplify \( -\frac{7\pi}{2} \):

\[ -\frac{7\pi}{2} = -3\pi - \frac{\pi}{2} = -3\pi + \frac{\pi}{2} = -\frac{6\pi}{2} + \frac{\pi}{2} = -\frac{5\pi}{2} \]

This angle can be further simplified by adding \( 4\pi \):

\[ -\frac{5\pi}{2} + 4\pi = -\frac{5\pi}{2} + \frac{8\pi}{2} = \frac{3\pi}{2} \]

### Step 8: Calculate \( Z^{14} \)
Now substituting back into the expression for \( Z^{14} \):

\[ Z^{14} = 128 \left( \cos\left(\frac{3\pi}{2}\right) + i \sin\left(\frac{3\pi}{2}\right) \right) \]

Using the values:

\[ \cos\left(\frac{3\pi}{2}\right) = 0, \quad \sin\left(\frac{3\pi}{2}\right) = -1 \]

Thus:

\[ Z^{14} = 128(0 - i) = -128i \]

### Final Answer
\[ Z^{14} = -128i \]