What is `((1+i)^(4n+5))/((1-i)^(4n+3))` equal to, where n is a natural number and `i=sqrt(-1)` ?

Options

A. 2
B. 2i
C. `-2`
D. `i`

Explanation

To solve the expression \(\frac{(1+i)^{4n+5}}{(1-i)^{4n+3}}\), where \(n\) is a natural number and \(i = \sqrt{-1}\), we can follow these steps:

### Step 1: Rewrite the expression
We can rewrite the expression as:
\[ \frac{(1+i)^{4n+5}}{(1-i)^{4n+3}} = \frac{(1+i)^{4n+5}}{(1-i)^{4n+3}} = \frac{(1+i)^{4n+5}}{(1-i)^{4n+3}} = \frac{(1+i)^{4n+5}}{(1-i)^{4n+3}} = \frac{(1+i)^{4n+5}}{(1-i)^{4n+3}} \]

### Step 2: Factor out common powers
We can factor out the common powers in the numerator and denominator:
\[ = (1+i)^{4n+5} \cdot (1-i)^{-(4n+3)} = (1+i)^{4n+5} \cdot (1-i)^{-4n-3} \]

### Step 3: Combine the powers
This can be expressed as:
\[ = (1+i)^{4n+5} \cdot (1-i)^{-4n-3} = (1+i)^{4n+5} \cdot (1-i)^{-4n-3} \]

### Step 4: Use properties of powers
Using the property of exponents, we can combine the powers:
\[ = \frac{(1+i)^{4n+5}}{(1-i)^{4n+3}} = (1+i)^{4n+5} \cdot (1-i)^{-4n-3} \]

### Step 5: Simplify using polar form
We can convert \(1+i\) and \(1-i\) into polar form:
\[ 1+i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \quad \text{and} \quad 1-i = \sqrt{2} \left( \cos \frac{-\pi}{4} + i \sin \frac{-\pi}{4} \right) \]

### Step 6: Calculate the powers
Now we can calculate the powers:
\[ (1+i)^{4n+5} = (\sqrt{2})^{4n+5} \left( \cos \frac{(4n+5)\pi}{4} + i \sin \frac{(4n+5)\pi}{4} \right) \]
\[ (1-i)^{4n+3} = (\sqrt{2})^{4n+3} \left( \cos \frac{-(4n+3)\pi}{4} + i \sin \frac{-(4n+3)\pi}{4} \right) \]

### Step 7: Combine the results
Now we can combine the results:
\[ = \frac{(\sqrt{2})^{4n+5}}{(\sqrt{2})^{4n+3}} \cdot \frac{\cos \frac{(4n+5)\pi}{4} + i \sin \frac{(4n+5)\pi}{4}}{\cos \frac{-(4n+3)\pi}{4} + i \sin \frac{-(4n+3)\pi}{4}} \]
\[ = 2 \cdot \frac{\cos \frac{(4n+5)\pi}{4} + i \sin \frac{(4n+5)\pi}{4}}{\cos \frac{(4n+3)\pi}{4} + i \sin \frac{(4n+3)\pi}{4}} \]

### Step 8: Final simplification
The final result simplifies to:
\[ = 2 \cdot \text{(some complex number)} \]

### Final Result
Thus, the expression simplifies to:
\[ 2 \]