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 proceed as follows:

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

We can multiply the numerator and denominator by the conjugate of the denominator:

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

Calculating this:

• The denominator becomes: $3^2 + 1^2 = 10.$
• The numerator is: $(-2)(3) + 2(-i)(3) - 2(-2i) + 2(-2i)(-i) = -6 + 2i + 2i - 2 = -8 + 4i.$

Thus: $\frac{z}{w} = \frac{-8 + 4i}{10} = -0.8 + 0.4i.$

To evaluate this integral, we will use integration by parts. Let:

• $u = (3x - 1)$, thus $du = 3dx$.
• $dv = \cos(\pi x)dx$, thus $v = \frac{1}{\pi}\sin(\pi x)$.

Applying integration by parts: $\int u \; dv = uv - \int v \; du.$

This gives us: $\int (3x - 1) \cos(\pi x) \; dx = (3x - 1) \cdot \frac{1}{\pi}\sin(\pi x) \bigg|_0^{\frac{1}{2}} - \int \frac{1}{\pi}\sin(\pi x) \cdot 3 \, dx.$

Evaluating the first part: At $x=\frac{1}{2}$: $(3(\frac{1}{2}) - 1) \cdot \frac{1}{\pi}\sin(\frac{\pi}{2}) = (1.5 - 1) \cdot \frac{1}{\pi}(1) = \frac{0.5}{\pi}.$ At $x=0$: $(3(0) - 1) \cdot \frac{1}{\pi}\sin(0) = 0.$

So that part contributes $\frac{0.5}{\pi}$.

Now, for: $- \int_0^{\frac{1}{2}} 3 \cdot \frac{1}{\pi}\sin(\pi x) \; dx = -\frac{3}{\pi} \left[-\frac{1}{\pi}\cos(\pi x) \right]_0^{\frac{1}{2}} = -\frac{3}{\pi^2} \left[-1 - 1\right] = -\frac{3}{\pi^2} (-2) = \frac{6}{\pi^2}.$

Thus the final result is: $\frac{0.5}{\pi} + \frac{6}{\pi^2}.$

To sketch this region, we first interpret the conditions provided:

• The condition $|z| \leq 2 - 2$ represents a circle centered at the origin with radius 2.
• The argument condition indicates that the angle of the vector $z$ lies between rac{\\pi}{4} and rac{\\pi}{4}.

This creates a wedge-shaped region in the Argand plane that is between the lines passing through the origin, at those angles. The intersection of this wedge with the circle gives the desired region.

To sketch the graph of: $y = \frac{x^2 - 1}{x^2},$ we can simplify this: $y = 1 - \frac{1}{x^2}.$

1. Intercepts:

   • x-intercepts: Set $y = 0$: $x^2 - 1 = 0 \implies x = \pm1.$
   • y-intercept: Set $x = 0$ (note it's undefined): The graph approaches $y = 1$ as $x \to \infty$ and $x \to -\infty$. Also, $y$ approaches negative infinity as $x \to 0$.

2. Sketch: Plot the intercepts and note the horizontal asymptote at $y = 1$. The graph will be in two parts, one above the x-axis and one below, creating a two-branch hyperbola-like shape. Indicate the critical points and asymptotes to complete the sketch.

First, we need to clarify the equation $y = 6 - y$, which represents a line. If we redefine it correctly: $y = 6 - x.$ The relevant line intersects the x-axis at $x = 6$.

To find the volume when this region is rotated about the x-axis, we can use the method of cylindrical shells: $V = 2\pi \int_0^{6} y \cdot h(x) \; dx = 2\pi \int_0^{6} (6 - x) \cdot x \; dx.$

Calculating:

1. Evaluate: $= 2\pi \int_0^{6} (6x - x^2) \; dx = 2\pi \left[ \frac{6x^2}{2} - \frac{x^3}{3} \right]_0^6$.
2. Compute: $= 2\pi \left[ 18 - 72\right] = 2\pi(-54) = -108\pi.$

Thus, the volume is: $V = 108\pi$. This positive value indicates the volume of the solid formed.