Show that the least positive integral value of n for which `((1+i)/(1 -i))^(n) = 1` is 4.

Answer

To solve the problem of finding the least positive integral value of \( n \) for which

\[ \left(\frac{1+i}{1-i}\right)^n = 1, \]

we will follow these steps:

### Step 1: Simplify the expression \(\frac{1+i}{1-i}\)

To simplify \(\frac{1+i}{1-i}\), we multiply the numerator and the denominator by the conjugate of the denominator, which is \(1+i\):

\[ \frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)(1+i)}{(1-i)(1+i)}. \]

### Step 2: Calculate the numerator and the denominator

Calculating the numerator:

\[ (1+i)(1+i) = 1 + 2i + i^2 = 1 + 2i - 1 = 2i. \]

Calculating the denominator:

\[ (1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2. \]

### Step 3: Combine the results

Now we can combine the results:

\[ \frac{1+i}{1-i} = \frac{2i}{2} = i. \]

### Step 4: Rewrite the equation

Now we rewrite the original equation:

\[ \left(i\right)^n = 1. \]

### Step 5: Determine the powers of \( i \)

Recall the powers of \( i \):

- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)

The powers of \( i \) repeat every 4 steps. Thus, \( i^n = 1 \) when \( n \) is a multiple of 4.

### Step 6: Find the least positive integral value of \( n \)

The least positive integral value of \( n \) that satisfies \( i^n = 1 \) is \( n = 4 \).

### Conclusion

Thus, we have shown that the least positive integral value of \( n \) for which

\[ \left(\frac{1+i}{1-i}\right)^n = 1 \]

is

\[ \boxed{4}. \]

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