Let $z = 2 + 3i$ and $w = 1 - i$ - HSC - SSCE Mathematics Extension 2 - Question 11 - 2018 - Paper 1

We first calculate $z - 2$:

$$z - 2 = (2 + 3i) - 2 = 3i.$$

Now, we compute:

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

Here we identify:

• $x = -\frac{3}{2}$ and $y = \frac{3}{2}$.

Given that $p(r) = 0$ and $p(4) = 0$ with $p(4)$ being a double root, we calculate:

1. The polynomial evaluated at $r$ gives: $$r^3 + ar^2 + b = 0.$$
2. The polynomial evaluated at the double root 4 yields: $$p(4) = 4^3 + 4a + b = 0 \implies 64 + 4a + b = 0.$$
3. Also, from the derivative $p'(x) = 3x^2 + 2ax$, we find: $$p'(4) = 3(4^2) + 2(4)a = 0 \implies 48 + 8a = 0 \implies a = -6.$$
4. Substituting $a = -6$ in $64 + 4(-6) + b = 0$ yields: $$64 - 24 + b = 0 \implies b = -40.$$
5. Lastly, substituting $b = -40$ in the original polynomial gives: $$p(r) = r^3 - 6r^2 - 40 = 0.$$

Using partial fraction decomposition, express:

$$\frac{x^2 - 6}{(x + 1)(x^2 - 3)} = \frac{a}{x + 1} + \frac{bx + c}{x^2 - 3}.$$

By equating coefficients:

1. Multiply both sides by ((x + 1)(x^2 - 3)
   • This gives a system of equations:
   • Solving will yield values for $a$, $b$, and $c$.
   • Finally, compute the integral using these expressions.