Find the integral values of n for the equations :
(a) `(1+i)^(n)=(1-i)^(n)`

Answer

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

### Step 1: Write the complex numbers in polar form
First, we need to express \(1+i\) and \(1-i\) in polar form.

The modulus of \(1+i\) is:
\[ |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2} \]
The argument of \(1+i\) is:
\[ \arg(1+i) = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \]
Thus, we can write:
\[ 1+i = \sqrt{2} \left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) = \sqrt{2} e^{i\frac{\pi}{4}} \]

For \(1-i\):
The modulus is the same:
\[ |1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \]
The argument of \(1-i\) is:
\[ \arg(1-i) = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \]
Thus, we can write:
\[ 1-i = \sqrt{2} \left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right) = \sqrt{2} e^{-i\frac{\pi}{4}} \]

### Step 2: Set up the equation
Now substituting these polar forms into the equation:
\[ (\sqrt{2} e^{i\frac{\pi}{4}})^{n} = (\sqrt{2} e^{-i\frac{\pi}{4}})^{n} \]
This simplifies to:
\[ (\sqrt{2})^{n} e^{i\frac{n\pi}{4}} = (\sqrt{2})^{n} e^{-i\frac{n\pi}{4}} \]

### Step 3: Cancel the common terms
Since \((\sqrt{2})^{n}\) is non-zero for any integer \(n\), we can divide both sides by \((\sqrt{2})^{n}\):
\[ e^{i\frac{n\pi}{4}} = e^{-i\frac{n\pi}{4}} \]

### Step 4: Solve for \(n\)
This implies:
\[ e^{i\frac{n\pi}{4}} = e^{-i\frac{n\pi}{4}} \implies e^{i\frac{n\pi}{4}} \cdot e^{i\frac{n\pi}{4}} = 1 \implies e^{i\frac{n\pi}{2}} = 1 \]
The equation \(e^{i\frac{n\pi}{2}} = 1\) holds true when:
\[ \frac{n\pi}{2} = 2k\pi \quad \text{for } k \in \mathbb{Z} \]
This simplifies to:
\[ n = 4k \quad \text{for } k \in \mathbb{Z} \]

### Step 5: Conclusion
Thus, the integral values of \(n\) that satisfy the equation are:
\[ n = 4k \quad \text{where } k \text{ is any integer.} \]