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

To find the reciprocal of a complex number, we use the formula:

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

To shade the region on the Argand diagram:

1. First inequality: The condition $0 \leq Re z \leq 2$ indicates that we should shade the area between the vertical lines $x = 0$ and $x = 2$.

2. Second inequality: The condition $|z - 1 + i| \leq 2$ describes a circle centered at $(1, 1)$ with a radius of $2$. We shade the area inside this circle.

Thus, the valid region is the overlapping area of these two shaded parts.

Since the coefficients of the polynomial $P(z)$ must be real, and given that $2 + i$ is a root, it follows that its conjugate $2 - i$ must also be a root.

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

where $r$ is the third root. Expanding the factors:

Thus:

Let $P(n)$ be the statement $(cos \theta - i sin \theta)^n = cos(n \theta) - i sin(n \theta).$

Base case ($n = 1$):

Inductive step: Assume true for $n = k$:

Then for $n = k + 1$:

Using product-to-sum identities, we can show:

Thus the formula holds for $n = k+1$.

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

Therefore, the simplified result is:

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

To find the real part, we compute:

Multiplying by the conjugate, we get:

From the previous calculation:

The imaginary part can be extracted as:

Thus, the imaginary part expressed in terms of $\theta$ is:

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