`((1 + i)^(2 n + 1))/( (1 - i) ^( 2 n - 1)), n in N` in `( r, theta )` form is

Options

A. (2,0)
B. `(2 , pi//2)`
C. `(2,pi)`
D. `(2, pi//4)`

Answer

To express the given expression \(\frac{(1 + i)^{2n + 1}}{(1 - i)^{2n - 1}}\) in the polar form \((r, \theta)\), we will follow these steps:

### Step 1: Convert \(1 + i\) and \(1 - i\) to Polar Form
First, we need to find the polar forms of \(1 + i\) and \(1 - i\).

1. **Magnitude of \(1 + i\)**:
\[ |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2} \]

2. **Argument of \(1 + i\)**:
\[ \theta_1 = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \]

Thus, the polar form of \(1 + i\) is:
\[ 1 + i = \sqrt{2} \left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) \]

3. **Magnitude of \(1 - i\)**:
\[ |1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \]

4. **Argument of \(1 - i\)**:
\[ \theta_2 = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \]

Thus, the polar form of \(1 - i\) is:
\[ 1 - i = \sqrt{2} \left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right) \]

### Step 2: Raise to the Powers
Now we raise both expressions to their respective powers.

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

2. **Denominator**:
\[ (1 - i)^{2n - 1} = \left(\sqrt{2}\right)^{2n - 1} \left(\cos\left((2n - 1)\left(-\frac{\pi}{4}\right)\right) + i\sin\left((2n - 1)\left(-\frac{\pi}{4}\right)\right)\right) \]
\[ = 2^{(2n - 1)/2} \left(\cos\left(-(2n - 1)\frac{\pi}{4}\right) + i\sin\left(-(2n - 1)\frac{\pi}{4}\right)\right) \]

### Step 3: Combine the Results
Now we can combine the results from the numerator and denominator:

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

### Step 4: Simplify Magnitudes and Angles
1. **Magnitude**:
\[ = 2^{\frac{(2n + 1) - (2n - 1)}{2}} = 2^{\frac{2}{2}} = 2 \]

2. **Angle**:
\[ = \left((2n + 1)\frac{\pi}{4} - (-(2n - 1)\frac{\pi}{4})\right) = \left((2n + 1 + 2n - 1)\frac{\pi}{4}\right) = \left(4n\frac{\pi}{4}\right) = n\pi \]

### Final Result
Thus, the expression in polar form \((r, \theta)\) is:
\[ (2, n\pi) \]