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:

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

This gives us: $= \frac{1 - i}{1 + i - i^2}$ $= \frac{1 - i}{1 + 1} = \frac{1 - i}{2}$

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

To shade the region defined by these inequalities:

1. For $0 \leq Re z \leq 2$:

   • This represents a vertical strip on the Argand plane from $x = 0$ to $x = 2$.

2. For $|z - 1 + i| \leq 2$:

   • This represents all points within or on the circle centered at $(1, 1)$ with radius 2.
   • The equation of this circle is given by $(x - 1)^2 + (y - 1)^2 \leq 4$.

Now, shade the area that overlaps between the vertical strip and the circle.

Since $P(z)$ has real coefficients, any non-real root must have its complex conjugate as a root as well. Given that $2 + i$ is a root, its conjugate $2 - i$ must also be a root.

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

1. Start with the factors corresponding to these roots: $(z - (2 + i))(z - (2 - i)) = (z - 2 - i)(z - 2 + i)$ Simplifying, this gives: $(z - 2)^2 + 1 = z^2 - 4z + 5$

2. Therefore, we can express $P(z)$ as: $P(z) = (z^2 - 4z + 5)(z + r) = z^3 + r z^2 + s z + 20$

Utilizing Vieta’s formulas, we can find $r$ and $s$ by matching coefficients.

To prove by induction:

Base case (n=1): $\cos \theta - i \sin \theta = \cos(1\theta) - i\sin(1\theta)$

• This is true.

Inductive step: Assume true for $n=k$, i.e., $(\cos \theta - i \sin \theta)^k = \cos(k\theta) - i\sin(k\theta)$ Then for $n=k+1$:

Given $z = 2(\cos \theta + i \sin \theta)$, we can find the conjugate as follows:

$\bar{z} = 2(\cos \theta - i \sin \theta)$

This defines the conjugate of a complex number given its polar form.

To find the real part:

1. First, calculate $\frac{1}{1 - z}$: $= \frac{1}{1 - 2(\cos \theta + i \sin \theta)} = \frac{1}{1 - 2 \cos \theta - 2i \sin \theta}$

2. Multiply numerator and denominator by the conjugate of the denominator: $= \frac{1(1 - 2 \cos \theta + 2i \sin \theta)}{(1 - 2 \cos \theta)^2 + (2 \sin \theta)^2}$

3. Simplifying the denominator: $= \frac{1 - 2 \cos \theta + 2i \sin \theta}{1 - 4 \cos \theta + 4 \cos^2 \theta + 4 \sin^2 \theta}$ Note that $\cos^2 \theta + \sin^2 \theta = 1$. Hence, $= \frac{1 - 2 \cos \theta + 2i \sin \theta}{5 - 4 \cos \theta}$

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

Building upon the previous steps, the imaginary part of $\frac{1}{1 - z}$ is:

1. From our earlier expression: $\frac{1 - 2 \cos \theta + 2i \sin \theta}{5 - 4 \cos \theta}$

2. The imaginary part is given by: $\text{Imaginary part} = \frac{2 \sin \theta}{5 - 4 \cos \theta}$

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