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

We start with the given complex numbers:

$z = -2 - 2i, \, w = 3 + i.$

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

$\frac{z}{w} = \frac{-2 - 2i}{3 + i} \cdot \frac{3 - i}{3 - i} = \frac{(-2)(3) + (-2)(-i) + (-2i)(3) + (-2i)(-i)}{9 + 1} = \frac{-6 + 2i - 6i - 2}{10} = \frac{-8 - 4i}{10} = -0.8 - 0.4i.$

Thus, the expression in the form $x + iy$ is:

$-0.8 - 0.4i.$

We will solve this integral using integration by parts.

Let:

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

Then we have:

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

Now applying integration by parts:

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

Substituting the values, we get:

$= \left(3x - 1\right) \left(\frac{1}{\pi} \sin(\pi x)\right) \Bigg|_{0}^{1} - \int_{0}^{1} \frac{1}{\pi} \sin(\pi x) (3 \, dx).$

Calculating the first term[= \left(3(1) - 1\right) \left(\frac{1}{\pi} \sin(\pi)\right) - \left(3(0) - 1\right) \left(\frac{1}{\pi} \sin(0)\right)= 0. ]

Now, evaluating the remaining integral: $- \frac{3}{\pi} \left[-\frac{1}{\pi} \cos(\pi x) \right]_{0}^{1} = -\frac{3}{\pi} \left[-\frac{1}{\pi} (-1 - 1)\right] = \frac{3}{\pi^{2}} (2) = \frac{6}{\pi^{2}}.$

Thus, $\int_{0}^{1} (3x - 1) \cos(\pi x) \, dx = \frac{6}{\pi^{2}}.$

To sketch the region:

1. Calculate the modulus $|2 - 2i|$: $|2 - 2i| = \sqrt{(2)^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}.$

2. The inequality $|z| \leq 2\sqrt{2}$ indicates a circle of radius $2\sqrt{2}$ centered at the origin.

3. The arguments $\arg z = \frac{\pi}{4}$ gives the line angled at $45^{\circ}$ above the real axis, and $\arg z = \frac{\pi}{4}$ is a duplicate line, indicating no area between these two angles. Thus, the region is:

   • The inside of a circle with radius $2\sqrt{2}$ restricted to the line at $\frac{\pi}{4}$.

To sketch this graph, we first find the intercepts:

1. x-intercepts: Set the numerator to zero: $x^{2} = 0 \Rightarrow x = 0.$ Thus, the x-intercept is at $(0, 0)$.

2. y-intercepts: Set $x=0$: $y = \frac{0^{2}}{1 - 0^{2}} = 0.$ So, the y-intercept is also at $(0, 0)$.

3. Identify vertical asymptotes where $1 - x^{2} = 0$: $1 - x^{2} = 0 \Rightarrow x = \pm 1.$ Thus, the graph has vertical asymptotes at $x = 1$ and $x = -1$.

4. Behavior around asymptotes: The function approaches infinity as it gets close to these points.

5. Draw the graph: The graph is symmetric around the y-axis since it is an even function. It rises steeply to infinity around the asymptotes at $x=1$ and $x=-1$.

To find the volume of the solid formed by rotating the region around the x-axis:

1. The volume $V$ using the method of cylindrical shells is given by: $V = 2\pi \int_{a}^{b} (radius)(height) \, dx.$

2. Setting

   • radius: Distance from the axis of rotation to the shell, which is $y$ as we rotate around the x-axis.
   • height: Given by $y = 6 - y$. Thus, $h = 6 - y$.

3. Determine limits of integration:

   • For $y$ from $0$ to $6$, so limits are $0$ to $6$.

4. Substitute into the volume equation: $V = 2\pi \int_{0}^{6} (x)(6 - y) \, dy.$

5. Solving this integral will yield the total volume of the solid.