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

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

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

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

To shade the region on the Argand diagram, we first interpret the inequalities:

1. The inequality $0 \leq Re z < 2$ means we are looking at the vertical strip between the lines $Re(z) = 0$ and $Re(z) = 2$.

2. The inequality $|z - 1 + i| < 2$ describes a circle centered at $(1,1)$ with a radius of 2.

The region where both inequalities hold is the area inside the circle that also lies within the vertical strip.

Since $P(z)$ has real coefficients, any non-real root must occur in conjugate pairs. Hence, if $2 + i$ is a root, its conjugate $2 - i$ must also be a root.

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

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

First, we find the quadratic factor:

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

Therefore, we can express $P(z)$ as:

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

We can find $r$ once the third root is known.

To prove by induction:

Base Case: For $n = 1$: $(cos \theta - i sin \theta)^1 = cos(1 \theta) - i sin(1 \theta)$ This holds true.

Inductive Step: Assume true for $n=k$: $(cos \theta - i sin \theta)^k = cos(k \theta) - i sin(k \theta)$ Now, show 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 product-to-sum identities: $= cos \theta cos(k \theta) + sin \theta sin(k \theta) - i (sin \theta cos(k \theta) - cos \theta sin(k \theta))$

This leads to: $= cos((k+1) \theta) - i sin((k+1) \theta)$

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

To find $\frac{1}{z}$: $z = 2(cos \theta + i sin \theta)$ We can use the polar form and calculate: $\frac{1}{z} = \frac{1}{2}(cos(-\theta) + i sin(-\theta))$ Thus, the answer is $\frac{1}{2} (cos(-\theta) + i sin(-\theta))$.

Using the formula for the real part, we compute:

$\frac{1}{1 - z} = \frac{1}{1 - 2(cos \theta + i sin \theta)}$ We simplify this to: $= \frac{1 - 2(cos \theta + i sin \theta)(1 - z)}{(1 - 2 cos \theta)^2 + (-2 sin \theta)^2}$ After resolving fractions, we ascertain:

$Re(\frac{1}{1-z}) = \frac{1 - 2 cos \theta}{5 - 4 cos^2 \theta}$.

For the imaginary part, we look at: $Im(\frac{1}{1 - z}) = \frac{-2 sin \theta}{(1 - 2 cos \theta)^2 + (2 sin \theta)^2}$ This simplifies to yield: $= \frac{-2 sin \theta}{5 - 4 cos^2 \theta}$. Thus, we express the imaginary part in terms of $\theta$.