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

To find $\frac{1}{w}$, we need to multiply by the conjugate:

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

Thus, the result is $\frac{1}{2} - \frac{1}{2}i$.

The inequality $0 \leq \text{Re} z \leq 2$ represents a vertical strip from $x=0$ to $x=2$ on the Argand plane. The inequality $|z - 1 + i| \leq 2$ represents a circle centered at $(1, 1)$ with a radius of 2.

To shade the region where both inequalities hold, we shade the strip and also the area inside the circle. The overlapping region of these two will be shaded.

Since the coefficients of the polynomial $P(z)$ are real numbers, the roots of the polynomial occur in conjugate pairs. Therefore, if $2 + i$ is a root, $2 - i$ must also be a root.

Given the roots $2 + i$ and $2 - i$, we can express the polynomial $P(z)$ as: $P(z) = (z - (2+i))(z - (2-i))(z - r)$

First, multiplying the conjugate pairs gives: $(z - (2+i))(z - (2-i)) = (z - 2 - i)(z - 2 + i) = (z - 2)^2 + 1 = z^2 - 4z + 5.$

Now, we need to add the factor corresponding to the third root $r$, which can be expressed as: $P(z) = (z^2 - 4z + 5)(z - r).$

We will use mathematical induction.

Base case: For $n = 1$, it holds that: $\left( \cos \theta - i \sin \theta \right)^1 = \cos \theta - i \sin \theta.$

Inductive step: Assume the formula holds for $n = k$, i.e., $\left( \cos \theta - i \sin \theta \right)^k = \cos(k\theta) - i \sin(k\theta).$

Then for $n = k + 1$: $\left( \cos \theta - i \sin \theta \right)^{k+1} = \left( \cos \theta - i \sin \theta \right) \left( \cos(k\theta) - i \sin(k\theta) \right).$

Expanding and rearranging using angle addition formulas, we have: $= \cos \theta \cos(k\theta) + \sin \theta \sin(k\theta) - i (\sin \theta \cos(k\theta) - \cos \theta \sin(k\theta))$ $= \cos((k + 1)\theta) - i \sin((k + 1)\theta).$ Thus, by induction, the formula holds for all integers $n \geq 1$.

Given $z = 2 \left( \cos \theta + i \sin \theta \right)$, $\frac{1}{z} = \frac{1}{2 \left( \cos \theta + i \sin \theta \right)} = \frac{1}{2} \cdot \frac{\cos \theta - i \sin \theta}{\cos^2 \theta + \sin^2 \theta} = \frac{1}{2} \left( \cos \theta - i \sin \theta \right).$

To find this, we first calculate: $\frac{1}{1 - z} = \frac{1}{1 - 2(\cos \theta + i \sin \theta)} = \frac{1}{1 - 2\cos \theta - 2i \sin \theta}.$ Multiplying by the conjugate: $\frac{1 - 2\cos \theta + 2i \sin \theta}{(1 - 2\cos \theta)^2 + (2\sin \theta)^2} = \frac{1 - 2\cos \theta + 2i \sin \theta}{1 - 4\cos \theta + 4\cos^2 \theta + 4\sin^2 \theta} = \frac{1 - 2\cos \theta + 2i \sin \theta}{(1 - 4\cos \theta + 4) = (5 - 4\cos \theta)}.$ Thus, the real part is $\frac{1 - 2\cos \theta}{5 - 4\cos \theta}$.

From the result in (ii), the imaginary part of $\frac{1}{1 - z}$ is: $\frac{2\sin \theta}{(5 - 4\cos \theta)}.$ Thus, the imaginary part expressed in terms of $\theta$ is $\frac{2\sin \theta}{5 - 4\cos \theta}$.