Let $z = 1 + 2i$ and $w = 1 + i$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2002 - Paper 1

$\frac{1}{w} = \frac{1}{1 + i} \cdot \frac{1 - i}{1 - i} = \frac{1 - i}{2} = \frac{1}{2} - \frac{1}{2}i$

The first inequality $0 \leq \text{Re} z \leq 2$ represents a vertical strip between the lines $\text{Re} z = 0$ and $\text{Re} z = 2$. The second inequality $|z - (1 + i)| \leq 2$ represents a circle centered at $(1, 1)$ with a radius of 2. The shaded region is the intersection of the strip and the circle.

Since $P(z)$ has real coefficients and $2 + i$ is a root, its conjugate $2 - i$ must also be a root.

Given the roots $2 + i$ and $2 - i$, we can factor them out: $P(z) = (z - (2 + i))(z - (2 - i))(z - r)$, where $r$ is the remaining real root. Multiplying these factors gives us the required factorization.

Base case: For $n = 1$, $\left( \cos \theta - i \sin \theta \right)^1 = \cos \theta - i \sin \theta$, which holds true. Inductive step: Assume it holds for $n = k$. Then for $n = k + 1$, we have: $\left( \cos \theta - i \sin \theta \right)^{k+1} = \left( \cos \theta - i \sin \theta \right)(\cos(k\theta) - i \sin(k\theta)) = \cos((k+1)\theta) - i \sin((k+1)\theta)$ by applying the angle addition formulas.

$\frac{1}{z} = \frac{1}{2 (\cos \theta + i \sin \theta)} = \frac{1}{2} (\cos(-\theta) - i \sin(-\theta)) = \frac{1}{2}\left( \cos \theta - i \sin \theta \right)$

To find $\frac{1}{1 - z}$ where $z = 2(\cos \theta + i \sin \theta)$, we first write it as $\frac{1}{1 - 2(\cos \theta + i \sin \theta)}$. Multiplying by the conjugate gives us the expression, which simplifies to show that the real part is indeed $\frac{1 - 2 \cos \theta}{5 - 4 \cos \theta}$.

Using the previous step, after simplification, we find the imaginary part of $\frac{1}{1 - z}$ as $\frac{2 \sin \theta}{5 - 4 \cos \theta}$.