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

Options

A. 16
B. 8
C. 4
D. 2

Answer

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 will go through the following steps:

### Step 1: Simplify the expression
We start with the expression:
\[ \frac{2i}{1+i} \]
To simplify this, we multiply the numerator and the denominator by the conjugate of the denominator, \( 1-i \):
\[ \frac{2i(1-i)}{(1+i)(1-i)} = \frac{2i(1-i)}{1^2 - i^2} = \frac{2i(1-i)}{1 - (-1)} = \frac{2i(1-i)}{2} \]
This simplifies to:
\[ i(1-i) = i - i^2 \]
Since \( i^2 = -1 \), we have:
\[ i - (-1) = i + 1 \]
Thus, we can rewrite the expression as:
\[ \frac{2i}{1+i} = i + 1 \]

### Step 2: Raise the expression to the power of \( n \)
Now we need to consider:
\[ (i + 1)^{n} \]

### Step 3: Determine when \( (i + 1)^{n} \) is a positive integer
Next, we need to find the smallest \( n \) such that \( (i + 1)^{n} \) is a positive integer.

### Step 4: Calculate \( (i + 1)^{n} \) for different values of \( n \)
1. For \( n = 1 \):
\[ (i + 1)^{1} = i + 1 \quad \text{(not a positive integer)} \]
2. For \( n = 2 \):
\[ (i + 1)^{2} = (i + 1)(i + 1) = i^2 + 2i + 1 = -1 + 2i + 1 = 2i \quad \text{(not a positive integer)} \]
3. For \( n = 3 \):
\[ (i + 1)^{3} = (i + 1)(2i) = 2i^2 + 2i = 2(-1) + 2i = -2 + 2i \quad \text{(not a positive integer)} \]
4. For \( n = 4 \):
\[ (i + 1)^{4} = (i + 1)^{2} \cdot (i + 1)^{2} = (2i)(2i) = 4i^2 = 4(-1) = -4 \quad \text{(not a positive integer)} \]
5. For \( n = 5 \):
\[ (i + 1)^{5} = (i + 1)^{4} \cdot (i + 1) = -4(i + 1) = -4i - 4 \quad \text{(not a positive integer)} \]
6. For \( n = 6 \):
\[ (i + 1)^{6} = (i + 1)^{4} \cdot (i + 1)^{2} = -4(2i) = -8i \quad \text{(not a positive integer)} \]
7. For \( n = 7 \):
\[ (i + 1)^{7} = (i + 1)^{4} \cdot (i + 1)^{3} = -4(-2 + 2i) = 8 - 8i \quad \text{(not a positive integer)} \]
8. For \( n = 8 \):
\[ (i + 1)^{8} = ((i + 1)^{4})^{2} = (-4)^{2} = 16 \quad \text{(this is a positive integer)} \]

### Conclusion
The least positive integer \( n \) such that \( \left(\frac{2i}{1+i}\right)^{n} \) is a positive integer is:
\[ \boxed{8} \]