The smallest integer n for which $(\frac{1+i}{1-i})^{n}=1$ is:

Q: The smallest integer n for which $(\frac{1+i}{1-i})^{n}=1$ is:

To determine the smallest integer $ n $ for which $\left(\frac{1+i}{1-i}\right)^{n} = 1$, we need to analyze the expression $\frac{1+i}{1-i}$. First, let's simplify $\frac{1+i}{1-i}$: 1. Multiply the numerator and the denominator by the conjugate of the denominator: \[ \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{(1-i)(1+i)} \] 2. Calculate the denominator: \[ (1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2 \] 3. Calculate the numerator: \[ (1+i)^2 = 1^2 + 2 \cdot 1 \cdot i + i^2 = 1 + 2i - 1 = 2i \] 4. Therefore, the expression simplifies to: \[ \frac{2i}{2} = i \] Now, we need to find the smallest integer $ n $ such that $ i^n = 1 $. The powers of $ i $ cycle every 4: - $ i^1 = i $ - $ i^2 = -1 $ - $ i^3 = -i $ - $ i^4 = 1 $ Thus, $ i^4 = 1 $. Therefore, the smallest integer $ n $ for which $ i^n = 1 $ is $ n = 4 $. **Explanation for why Option 1 is correct:** - The expression $\frac{1+i}{1-i}$ simplifies to $ i $. - The powers of $ i $ repeat every 4, with $ i^4 = 1 $. - Therefore, the smallest $ n $ such that $\left(\frac{1+i}{1-i}\right)^{n} = 1$ is $ n = 4 $. **Explanation for why other options are incorrect:** - **Option 2: $ n = 5 $**: If $ n = 5 $, then $ i^5 = i \neq 1 $. Therefore, $ n = 5 $ does not satisfy the condition. - **Option 3: $ n = 1 $**: If $ n = 1 $, then $ i^1 = i \neq 1 $. Therefore, $ n = 1 $ does not satisfy the condition. - **Option 4: $ n = 3 $**: If $ n = 3 $, then $ i^3 = -i \neq 1 $. Therefore, $ n = 3 $ does not satisfy the condition. Thus, Option 1 is the correct answer, as $ n = 4 $ is the smallest integer for which $\left(\frac{1+i}{1-i}\right)^{n} = 1$.

Question

Accepted Answer

To determine the smallest integer $ n $ for which $\left(\frac{1+i}{1-i}\right)^{n} = 1$, we need to analyze the expression $\frac{1+i}{1-i}$.

First, let's simplify $\frac{1+i}{1-i}$:

1. Multiply the numerator and the denominator by the conjugate of the denominator:
   $$
   \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{(1-i)(1+i)}
   $$

2. Calculate the denominator:
   $$
   (1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2
   $$

3. Calculate the numerator:
   $$
   (1+i)^2 = 1^2 + 2 \cdot 1 \cdot i + i^2 = 1 + 2i - 1 = 2i
   $$

4. Therefore, the expression simplifies to:
   $$
   \frac{2i}{2} = i
   $$

Now, we need to find the smallest integer $ n $ such that $ i^n = 1 $.

The powers of $ i $ cycle every 4:
- $ i^1 = i $
- $ i^2 = -1 $
- $ i^3 = -i $
- $ i^4 = 1 $

Thus, $ i^4 = 1 $. Therefore, the smallest integer $ n $ for which $ i^n = 1 $ is $ n = 4 $.

**Explanation for why Option 1 is correct:**

- The expression $\frac{1+i}{1-i}$ simplifies to $ i $.
- The powers of $ i $ repeat every 4, with $ i^4 = 1 $.
- Therefore, the smallest $ n $ such that $\left(\frac{1+i}{1-i}\right)^{n} = 1$ is $ n = 4 $.

**Explanation for why other options are incorrect:**

- **Option 2: $ n = 5 $**: If $ n = 5 $, then $ i^5 = i \neq 1 $. Therefore, $ n = 5 $ does not satisfy the condition.
- **Option 3: $ n = 1 $**: If $ n = 1 $, then $ i^1 = i \neq 1 $. Therefore, $ n = 1 $ does not satisfy the condition.
- **Option 4: $ n = 3 $**: If $ n = 3 $, then $ i^3 = -i \neq 1 $. Therefore, $ n = 3 $ does not satisfy the condition.

Thus, Option 1 is the correct answer, as $ n = 4 $ is the smallest integer for which $\left(\frac{1+i}{1-i}\right)^{n} = 1$.