If `((2+2i)/(2-2i))^(n)=1` find the least positive integral of n.

Answer

To solve the equation \(\left(\frac{2 + 2i}{2 - 2i}\right)^n = 1\), we will follow these steps:

### Step 1: Simplify the expression
We start by simplifying the fraction \(\frac{2 + 2i}{2 - 2i}\). We can factor out 2 from both the numerator and the denominator:

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

### Step 2: Rationalize the denominator
Next, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator:

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

### Step 3: Calculate the numerator and denominator
Now, we calculate the numerator and denominator separately:

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

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

Putting it all together, we have:

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

### Step 4: Substitute back into the equation
Now we substitute back into our original equation:

\[ (i)^n = 1 \]

### Step 5: Determine when \(i^n = 1\)
The powers of \(i\) cycle every 4:

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

Thus, \(i^n = 1\) when \(n\) is a multiple of 4. The least positive integer \(n\) that satisfies this condition is:

\[ n = 4 \]

### Final Answer
The least positive integral value of \(n\) is \(4\).

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