If `z_(1)` and `z_(2)` are two of the `8^(th)` roots of unity such that arg`(z_(1)/z_(2))` is last positive, then `z_(1)/z_(2)` is

Options

A. `1+i`
B. `1-i`
C. `(1+i)/sqrt(2)`
D. `(1-i)/sqrt(2)`

Explanation

To solve the problem, we need to find the expression for \( \frac{z_1}{z_2} \) where \( z_1 \) and \( z_2 \) are two of the 8th roots of unity, and the argument \( \text{arg}\left(\frac{z_1}{z_2}\right) \) is the last positive value.

### Step-by-Step Solution:

1. **Identify the 8th Roots of Unity**:
The 8th roots of unity can be expressed as:
\[ z_k = e^{i \frac{2k\pi}{8}} = e^{i \frac{k\pi}{4}} \quad \text{for } k = 0, 1, 2, \ldots, 7 \]
This gives us the roots:
\[ z_0 = e^{i \cdot 0} = 1, \quad z_1 = e^{i \frac{\pi}{4}}, \quad z_2 = e^{i \frac{\pi}{2}}, \quad z_3 = e^{i \frac{3\pi}{4}}, \quad z_4 = e^{i \pi}, \quad z_5 = e^{i \frac{5\pi}{4}}, \quad z_6 = e^{i \frac{3\pi}{2}}, \quad z_7 = e^{i \frac{7\pi}{4}} \]

2. **Express \( z_1 \) and \( z_2 \)**:
Let’s denote \( z_1 = z_k = e^{i \frac{k\pi}{4}} \) and \( z_2 = z_{k-1} = e^{i \frac{(k-1)\pi}{4}} \) for some integer \( k \).

3. **Calculate \( \frac{z_1}{z_2} \)**:
Now, we can find \( \frac{z_1}{z_2} \):
\[ \frac{z_1}{z_2} = \frac{e^{i \frac{k\pi}{4}}}{e^{i \frac{(k-1)\pi}{4}}} = e^{i \left(\frac{k\pi}{4} - \frac{(k-1)\pi}{4}\right)} = e^{i \frac{\pi}{4}} \]

4. **Determine the Argument**:
The argument of \( \frac{z_1}{z_2} \) is:
\[ \text{arg}\left(\frac{z_1}{z_2}\right) = \frac{\pi}{4} \]

5. **Identify the Last Positive Argument**:
Since we are looking for the last positive argument, we consider the next consecutive roots. If \( z_1 \) and \( z_2 \) are consecutive roots, the argument \( \text{arg}\left(\frac{z_1}{z_2}\right) \) will be the smallest positive angle between them, which is \( \frac{\pi}{4} \).

6. **Final Expression**:
Thus, we conclude that:
\[ \frac{z_1}{z_2} = e^{i \frac{\pi}{4}} = \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \]

### Final Answer:
\[ \frac{z_1}{z_2} = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \]