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 first find the modulus:

$$|w| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2.$$

Next, we find the argument, which is given by:

$$\arg(w) = \tan^{-1}\left(\frac{\Im(w)}{\Re(w)}\right) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}.$$

Therefore, we can express $w$ as:

$$w = 2 \text{ cis } \frac{\pi}{3}.$$

Using De Moivre's Theorem, we can compute:

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

Since $\text{ cis } 8\pi = 1$ (because it corresponds to a full circle), we have:

$$w^{24} = 2^{24} imes 1 = 16777216.$$

To find $A$, $B$, and $C$, we start by equating:

$$x^2 + 8x + 11 = A(x^2 + 2) + (Bx + C)(x - 3).$$

Expanding the right side:

$$Ax^2 + 2A + Bx^2 - 3Bx + Cx - 3C.$$

Combining like terms:

$$(A + B)x^2 + (-3B + C)x + (2A - 3C).$$

Now we equate coefficients:

1. For $x^2$: $A + B = 1$
2. For $x$: $-3B + C = 8$
3. For constant term: $2A - 3C = 11$.

Solving this system will yield values for $A, B,$ and $C$.

To factorise the expression, we start by using the quadratic formula:

$$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$

where $a = 1$, $b = 4i$, and $c = 5$.

Calculating the discriminant:

$$b^2 - 4ac = (4i)^2 - 4(1)(5) = -16 - 20 = -36.$$

Now substituting into the quadratic formula:

$$z = \frac{-4i \pm \sqrt{-36}}{2} = \frac{-4i \pm 6i}{2}.$$

This gives us:

$$z = \frac{-4i + 6i}{2}, \quad \frac{-4i - 6i}{2} = i, -5i.$$

Thus, we can factor as:

$$(z - i)(z + 5i).$$

To evaluate the integral, we can use the substitution:

Let $x = \sin(\theta)$, then $dx = \cos(\theta) d\theta$. The limits change from $0$ to $\frac{\pi}{2}$.

The integral becomes:

$$\int_0^{\frac{\pi}{2}} \sin^3(\theta) \sqrt{1 - \sin^2(\theta)} \cos(\theta) d\theta = \int_0^{\frac{\pi}{2}} \sin^3(\theta) \cos^2(\theta) d\theta.$$

Using the identity: $\cos^2(\theta) = 1 - \sin^2(\theta)$, we can rewrite:

$$\int_0^{\frac{\pi}{2}} \sin^3(\theta) (1 - \sin^2(\theta)) d\theta.$$

Evaluating the integral involves standard techniques such as integration by parts or further substitution.

To sketch the region defined by the inequality $z^2 + \bar{z}^2 \leq 8$, we start from the expression:

Let $z = x + iy$, then $\bar{z} = x - iy$. Thus:

$$z^2 = (x + iy)^2 = x^2 - y^2 + 2xyi, \quad \bar{z}^2 = (x - iy)^2 = x^2 - y^2 - 2xyi.$$

Adding these gives:

$$z^2 + \bar{z}^2 = 2(x^2 - y^2).$$

Setting this equal to $8$ gives:

$$2(x^2 - y^2) \ ext{ thus solving yields the boundary } x^2 - y^2 = 4.$$

Graphically, this represents a hyperbola in the Argand diagram. The region defined by the inequality will be the area within this boundary.