Find the least positive integer `n` such that `((2i)/(1+i))^n` is a positive integer.

Options

A. n =6
B. n =5
C. n =8
D. n =4

Explanation

To solve the problem of finding the least positive integer \( n \) such that \( \left( \frac{2i}{1+i} \right)^n \) is a positive integer, we can follow these steps:

### Step 1: Simplify \( \frac{2i}{1+i} \)

Multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{2i}{1+i} \cdot \frac{1-i}{1-i} = \frac{2i(1-i)}{(1+i)(1-i)} \]

### Step 2: Calculate the denominator

The denominator can be simplified as follows:

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

### Step 3: Calculate the numerator

Now calculate the numerator:

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

### Step 4: Combine the results

Now we can write:

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

### Step 5: Express \( (1+i)^n \)

Now we need to find \( (1+i)^n \) and determine when it is a positive integer.

### Step 6: Convert \( 1+i \) to polar form

To convert \( 1+i \) to polar form, we calculate its modulus and argument:

- Modulus:
\[ |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2} \]

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

Thus, we can express \( 1+i \) in polar form:

\[ 1+i = \sqrt{2} \left( \cos\frac{\pi}{4} + i \sin\frac{\pi}{4} \right) \]

### Step 7: Raise to the power of \( n \)

Now, raising \( 1+i \) to the power of \( n \):

\[ (1+i)^n = \left( \sqrt{2} \right)^n \left( \cos\left(\frac{n\pi}{4}\right) + i \sin\left(\frac{n\pi}{4}\right) \right) \]

### Step 8: Determine when it is a positive integer

For \( (1+i)^n \) to be a positive integer, the imaginary part must be zero and the real part must be positive. This occurs when:

1. \( \sin\left(\frac{n\pi}{4}\right) = 0 \) (which occurs for \( n = 4k \), where \( k \) is an integer)
2. \( \cos\left(\frac{n\pi}{4}\right) > 0 \)

### Step 9: Check values of \( n \)

- For \( n = 4 \):
\[ \cos\left(\frac{4\pi}{4}\right) = \cos(\pi) = -1 \quad \text{(not positive)} \]

- For \( n = 8 \):
\[ \cos\left(\frac{8\pi}{4}\right) = \cos(2\pi) = 1 \quad \text{(positive)} \]

### Conclusion

Thus, the least positive integer \( n \) such that \( \left( \frac{2i}{1+i} \right)^n \) is a positive integer is:

\[ \boxed{8} \]