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. `-2i`
D. i

Answer

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: Simplify the expression using properties of complex numbers
We start by rewriting the denominator:
\[ (1-i) = \frac{(1+i)(1-i)}{(1+i)} = \frac{1^2 - i^2}{(1+i)} = \frac{1 - (-1)}{(1+i)} = \frac{2}{(1+i)} \]
Thus, we can rewrite the expression:
\[ \frac{(1+i)^{4n+5}}{(1-i)^{4n+3}} = \frac{(1+i)^{4n+5}}{\left(\frac{2}{(1+i)}\right)^{4n+3}} = \frac{(1+i)^{4n+5} \cdot (1+i)^{4n+3}}{2^{4n+3}} \]

### Step 2: Combine the powers of \(1+i\)
Now we can combine the powers in the numerator:
\[ (1+i)^{(4n+5) + (4n+3)} = (1+i)^{8n+8} \]
So the expression becomes:
\[ \frac{(1+i)^{8n+8}}{2^{4n+3}} \]

### Step 3: Factor out the common terms
We can factor the exponent of \(1+i\):
\[ (1+i)^{8(n+1)} = ((1+i)^8)^{n+1} \]
Now we can rewrite the expression:
\[ \frac{((1+i)^8)^{n+1}}{2^{4n+3}} \]

### Step 4: Calculate \((1+i)^8\)
To find \((1+i)^8\), we first find \((1+i)^2\):
\[ (1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i \]
Then we find \((1+i)^4\):
\[ (1+i)^4 = (2i)^2 = 4i^2 = 4(-1) = -4 \]
Next, we find \((1+i)^8\):
\[ (1+i)^8 = (-4)^2 = 16 \]

### Step 5: Substitute back into the expression
Now we substitute back into our expression:
\[ \frac{(16)^{n+1}}{2^{4n+3}} = \frac{16^{n+1}}{2^{4n+3}} = \frac{(2^4)^{n+1}}{2^{4n+3}} = \frac{2^{4(n+1)}}{2^{4n+3}} = 2^{4(n+1) - (4n+3)} = 2^{4n + 4 - 4n - 3} = 2^{1} = 2 \]

### Final Answer
Thus, the value of \(\frac{(1+i)^{4n+5}}{(1-i)^{4n+3}}\) is equal to \(2\).

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