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

To find $\frac{1}{w}$, where $w = 1 + i$, we multiply the numerator and the denominator by the conjugate of the denominator:

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

Simplifying gives:

$= \frac{1}{2} - \frac{1}{2}i$

Therefore, the answer is $\frac{1}{2} - \frac{1}{2}i$.

To shade the region, we start by analyzing the inequalities:

1. The inequality $0 \leq \text{Re} z \leq 2$ indicates that the shaded region is bounded vertically between real parts of 0 and 2.
2. The inequality $|z - (1 - i)| \leq 2$ describes a circle centered at $(1, 1)$ with a radius of 2.

The combined region will be the overlap of the vertical strip between Re(z) = 0 and Re(z) = 2 along with the circle defined by the second inequality. The region can be illustrated in the Argand diagram by shading the portion of the circle that lies between the vertical lines at Re(z) = 0 and Re(z) = 2.

The polynomial $P(z)$ has real coefficients, which implies that any complex root must have its conjugate also as a root. Since $2 + i$ is a root, its conjugate, $2 - i$, must also be a root of $P(z)$.

Given that $2 + i$ and $2 - i$ are roots, we can express $P(z)$ as:

$P(z) = (z - (2 + i))(z - (2 - i))(z - r)$

Calculating the product of the first two factors:

$= (z - 2 - i)(z - 2 + i) = (z - 2)^2 + 1 = z^2 - 4z + 5$

Thus:

$P(z) = (z^2 - 4z + 5)(z - r)$

where $r$ is a real number determined from the constant term of the original polynomial.

To prove this by induction, we start with the base case where $n = 1$:

$\cos \theta - i \sin \theta = \cos(1\theta) - i \sin(1\theta).$

Now, assume that the formula holds for some integer $k$:

$(\cos \theta - i \sin \theta)^k = \cos(k\theta) - i \sin(k\theta).$

For $n = k + 1$:

i.e., we need to show:

$(\cos \theta - i \sin \theta)^{k+1} = (\cos \theta - i \sin \theta)(\cos(k\theta) - i \sin(k\theta)).$

Using the properties of complex multiplication:
$= \cos \theta \cos(k\theta) + \sin \theta \sin(k\theta) - i(\sin \theta \cos(k\theta) - \cos \theta \sin(k\theta)) = \cos((k + 1) \theta) - i \sin((k + 1) \theta).$

Thus, the induction holds and the statement is proved.

Given $z = 2 (\cos \theta + i \sin \theta)$:

First, find $1 - z = 1 - 2(\cos \theta + i \sin \theta) = 1 - 2 \cos \theta - 2i \sin \theta.$

To find $\frac{1}{1 - z}$, we multiply by the conjugate:

$\frac{1}{1 - z} = \frac{1 - z}{|1 - z|^2}$

Calculate $|1 - z|^2$:

$= (1 - 2 \cos \theta)^2 + (-2 \sin \theta)^2 = 1 - 4 \cos \theta + 4 \cos^2 \theta + 4 \sin^2 \theta = 1 - 4 \cos \theta + 4,$

using $\cos^2 \theta + \sin^2 \theta = 1$. Thus,

$|1 - z|^2 = 5 - 4 \cos \theta.$

So,

Therefore, the real part is: $\frac{1 - 2 \cos \theta}{5 - 4 \cos \theta}.$

From the previous computation, we have:

$\frac{1}{1 - z} = \frac{(1 - 2 \cos \theta) + 2i \sin \theta}{5 - 4 \cos \theta}.$

Thus, the imaginary part is:

$\frac{2 \sin \theta}{5 - 4 \cos \theta}.$

Therefore, the result is: $$\frac{2 \sin \theta}{5 - 4 \cos \theta}.$