Let $z = 2 - i oot{3}$ and $w = 1 + i oot{3}$ - HSC - SSCE Mathematics Extension 2 - Question 11 - 2013 - Paper 1

To express $w$ in modulus-argument form, we need to find the modulus and argument of $w = 1 + i\sqrt{3}$.

1. Modulus: $|w| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2.$
2. Argument: The argument can be found using:
   $\text{arg}(w) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}.$
   Thus, we can express $w$ in modulus-argument form as: $w = 2 \text{cis}\left(\frac{\pi}{3}\right).$

Using De Moivre's Theorem:

$w^{24} = \left(2 \text{cis}\left(\frac{\pi}{3}\right)\right)^{24} = 2^{24} \text{cis}\left(8\pi\right).$

Since $\text{cis}(8\pi) = 1$, we have:

$w^{24} = 2^{24} \cdot 1 = 2^{24}.$

To find $A$, $B$, and $C$, we will perform partial fraction decomposition on:

$\frac{x^2 + 8x + 11}{(x - 3)(x^2 + 2)}.\n$
Equating both sides leads us to:

$x^2 + 8x + 11 = A(x^2 + 2) + (Bx + C)(x - 3).$
After expanding and comparing coefficients, we can solve for $A$, $B$, and $C$.

We will use the quadratic formula:

$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 1$, $b = 4i$, and $c = 5$. We calculate:

$b^2 - 4ac = (4i)^2 - 4(1)(5) = -16 - 20 = -36.\n$
The roots are then:

$z = \frac{-4i \pm \sqrt{-36}}{2} = \frac{-4i \pm 6i}{2} = -2i \pm 3i.$
Thus, we can factorise:

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

We can use the substitution $u = x^2$, then $du = 2xdx$ or $dx = \frac{du}{2\sqrt{u}}$. This transforms the integral into:

$\int_0^1 x^3\sqrt{1 - x^2} dx = \frac{1}{2}\int_0^1 (u \cdot \sqrt{1 - u}) du.$
Solving this integral involves standard techniques and leads us to the result.

The inequality $z^2 + z^2 \leq 8$ simplifies to $2z^2 \leq 8$, or $z^2 \leq 4$.
This represents a circle of radius 2 in the Argand diagram centered at the origin. We sketch the circle and shade the interior to indicate all points that satisfy this inequality.