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