Simiplify : `((1+i)/(1-i))^(4n+1)` (n is a positive integer)

Answer

To simplify the expression \(\left(\frac{1+i}{1-i}\right)^{4n+1}\), we can follow these steps:

### Step 1: Rationalize the Denominator
We start with the expression \(\frac{1+i}{1-i}\). To simplify this, 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: Expand the Numerator and Denominator
Now we expand both the numerator and the denominator:

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

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

So, we have:

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

### Step 3: Substitute Back into the Original Expression
Now we substitute this result back into the original expression:

\[ \left(\frac{1+i}{1-i}\right)^{4n+1} = (i)^{4n+1} \]

### Step 4: Simplify \(i^{4n+1}\)
We know that the powers of \(i\) cycle every 4:
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)

To simplify \(i^{4n+1}\), we can express it as:

\[ i^{4n+1} = (i^4)^n \cdot i^1 = 1^n \cdot i = i \]

### Conclusion
Thus, the simplified form of the expression \(\left(\frac{1+i}{1-i}\right)^{4n+1}\) is:

\[ \boxed{i} \]