Consider the complex numbers $z = -2 - 2i$ and $w = 3 + i$ - HSC - SSCE Mathematics Extension 2 - Question 11 - 2014 - Paper 1

To find $\frac{z}{w}$, we compute:

$\frac{z}{w} = \frac{-2 - 2i}{3 + i}.$

To simplify, multiply the numerator and denominator by the conjugate of the denominator:

$\frac{-2 - 2i}{3 + i} \cdot \frac{3 - i}{3 - i} = \frac{(-2)(3) + 2(-2i) + 2i(3) - 2(-i^2)}{(3^2 + 1^2)}.$

This simplifies to:

$= \frac{-6 - 4i + 6i + 2}{10} = \frac{-4 + 2i}{10} = -\frac{2}{5} + \frac{1}{5}i.$

Thus, we have:

$\frac{z}{w} = -\frac{2}{5} + \frac{1}{5}i.$

To evaluate the integral, we need to use integration by parts. We identify:

• Let $u = 3x - 1$ and $dv = \cos(\pi x) \, dx$.

Calculating $du$ and $v$ gives:

• $du = 3 \, dx$;
• $v = \frac{1}{\pi} \sin(\pi x)$.

Applying integration by parts:

$\int u \, dv = uv - \int v \, du.$

Substituting in our values yields:

$\int (3x - 1) \cos(\pi x) \, dx = (3x - 1) \cdot \frac{1}{\pi} \sin(\pi x) \bigg|_0^1 - \int \frac{1}{\pi} \sin(\pi x) \cdot 3 \, dx.$

After evaluating the boundary terms and simplifying: $= 0 - \frac{3}{\pi} \left( -\frac{1}{\pi} \right) = \frac{3}{\pi^2}.$

This region in the Argand diagram represents the area within the circle of radius 2, centered at the origin.

The angles ( \frac{-\pi}{4} ) and ( \frac{\pi}{4} ) represent the boundaries of the region, forming a wedge in the first and fourth quadrants. Draw the circle and indicate the shaded area accordingly.

To sketch the graph, observe the function:

• Intercepts can be found by setting $y = 0$:

$x^2 - 1 = 0 \Rightarrow x = \pm 1.$

• The vertical asymptotes occur when the denominator is zero, which corresponds to points where $x^2 = 0$ (undefined)
• The graph approaches $y = 1$ as $x \to \infty$ and $y = -1$ as $x \to -\infty$. The overall shape resembles a hyperbola opening downward.

No calculus is needed; this will be a simple sketch of the characteristics.

To find the volume of the solid formed by revolving $y = 6 - y$ around the x-axis, we use the formula:

$V = 2\pi \int_a^b r(y)h(y) \, dy,$

where $r(y)$ is the distance from the axis of rotation to the curve and $h(y)$ is the height.

Here, the limits are determined by the intersection with the x-axis (set $y = 0$):

• So we find the volume: $= 2\pi \int_0^{6} (y)(6 - y) \, dy.$

Integrating will yield the volume of the solid.