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

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

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

Thus, $\frac{1}{w} = \frac{1}{2} - \frac{1}{2}i.$

1. The inequality $0 \leq Re z \leq 2$ describes a vertical strip on the Argand diagram from $Re z = 0$ to $Re z = 2$.

2. The inequality $|z - 1 + i| \leq 2$ describes a circle centered at the point $1 - i$ with a radius of 2.

3. Thus, the shaded region is the overlap of the vertical strip and the circle, indicating the points that satisfy both inequalities.

Since $P(z)$ has real coefficients, complex roots must occur in conjugate pairs. Therefore, if $2 + i$ is a root, $2 - i$ must also be a root.

We know $P(z)$ has roots $2 + i$ and $2 - i$. Thus, we can write:

$(z - (2+i))(z - (2-i)) = (z - 2 - i)(z - 2 + i) = (z - 2)^2 + 1.$

Now we can factor $P(z)$ by first finding the quadratic factor: $P(z) = (z - 2)^2 + 1 = z^2 - 4z + 5.$

To find the cubic factor, we must consider that the polynomial has one additional real root, say $k$. Thus the factorization would look like: $P(z) = (z - 2)^2 + 1(z - k),$ where $k$ must be determined based on additional information.

1. Base Case (n=1): $cos(\theta) - i sin(\theta) = cos(1\theta) - i sin(1\theta),$ which holds true.

2. Inductive Step: Assume the statement is true for some integer $k$, i.e., $(cos(\theta) - i sin(\theta))^k = cos(k\theta) - i sin(k\theta).$

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

Using the angle addition formulas: $= cos(\theta + k\theta) - i sin(\theta + k\theta) = cos((k + 1)\theta) - i sin((k + 1)\theta),$ which proves the statement for $n = k + 1$.

Thus, by induction the statement holds for all integers $n \geq 1$.

Given $z = 2(cos \theta + i sin \theta)$, we can find its reciprocal: $\frac{1}{z} = \frac{1}{2(cos \theta + i sin \theta)} = \frac{1}{2} \cdot \frac{cos(-\theta) - i sin(-\theta)}{cos^2(\theta) + sin^2(\theta)} = \frac{1}{2} \cdot (cos(-\theta) - i sin(-\theta)) = \frac{1}{2}(cos \theta - i sin \theta).$

To find the real part of $\frac{1}{1 - z}$, we begin with: $\frac{1}{1 - z} = \frac{1}{1 - 2(cos \theta + i sin \theta)} \cdot \frac{1 + 2(cos \theta - i sin \theta)}{1 + 2(cos \theta - i sin \theta)} = \frac{1 + 2cos \theta - 2is \theta}{1 - 4cos \theta + 4}.$

Simplifying the denominator gives: $\frac{1 + 2cos \theta - 2i sin \theta}{5 - 4cos \theta}.$ The real part is $\frac{1 + 2cos \theta}{5 - 4cos \theta}$. To match the assertion, further simplification is needed.

From the earlier expression: $\frac{1 + 2cos \theta - 2i sin \theta}{5 - 4cos \theta}.$ The imaginary part is given by: $\frac{-2sin \theta}{5 - 4cos \theta}.$ Thus, the final expression for the imaginary part of $\frac{1}{1 - z}$ in terms of $\theta$ is: $-\frac{2sin \theta}{5 - 4cos \theta}.$