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To solve problem **1(a)**, which is an algebraic expression in complex numbers, we need to evaluate:

\[
\frac{(1 + i)^{28}}{(1 - i)^{24} - i(1 + i)^{24}}
\]

where \( i \) is the imaginary unit such that \( i^2 = -1 \).

Let's break down the expression step by step.

### Step 1: Simplify \( (1 + i) \) and \( (1 - i) \)
1. **Magnitude of \(1 + i\):**
   \[
   |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}
   \]
   So, we can write \( 1 + i \) in polar form as:
   \[
   1 + i = \sqrt{2} \left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right)
   \]

2. **Magnitude of \(1 - i\):**
   \[
   |1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}
   \]
   And similarly:
   \[
   1 - i = \sqrt{2} \left(\cos \left(-\frac{\pi}{4}\right) + i \sin \left(-\frac{\pi}{4}\right)\right)
   \]

### Step 2: Use De Moivre's Theorem
Using De Moivre’s Theorem:
1. **Compute \( (1 + i)^{28} \):**
   \[
   (1 + i)^{28} = \left(\sqrt{2}\right)^{28} \left(\cos \frac{28\pi}{4} + i \sin \frac{28\pi}{4}\right)
   \]
   Since \( \frac{28\pi}{4} = 7\pi \), which is equivalent to \( \pi \) (since \( 7\pi \) is an odd multiple of \( \pi \)),
   \[
   (1 + i)^{28} = 2^{14} (\cos \pi + i \sin \pi) = 2^{14} (-1) = -2^{14}
   \]

2. **Compute \( (1 - i)^{24} \):**
   \[
   (1 - i)^{24} = \left(\sqrt{2}\right)^{24} \left(\cos \frac{-24\pi}{4} + i \sin \frac{-24\pi}{4}\right)
   \]
   Since \( \frac{-24\pi}{4} = -6\pi \), which is equivalent to \( 0 \) (since \( -6\pi \) is an even multiple of \( \pi \)),
   \[
   (1 - i)^{24} = 2^{12} (\cos 0 + i \sin 0) = 2^{12}
   \]

### Step 3: Substitute and Simplify
Now, substitute back into the expression:
\[
\frac{(1 + i)^{28}}{(1 - i)^{24} - i(1 + i)^{24}}
\]

Since \((1 + i)^{24} = 2^{12} (\cos \pi + i \sin \pi) = -2^{12}\), we can substitute

Solve number 1(a): \[ \frac{(1 + i)^{28}}{(1 - i)^{24} - i(1 + i)^{24}} \] where \( i \) is the imaginary unit such that \( i^2 = -1 \).

Step-by-step solution for evaluating the complex expression \((1 + i)^{28} / ((1 - i)^{24} - i(1 + i)^{24})\) using De Moivre's theorem and polar forms. This problem introduces complex exponentiation and is suitable for advanced high school or early undergraduate students.

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