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

We first calculate $\frac{z - 2}{w}$:

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

To express this in the form $x + iy$, we can multiply numerator and denominator by the conjugate of the denominator:

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

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

Since $p(x) = x^3 + ax^2 + b$ has a double zero at $x = 4$, we can denote the roots as $r$, $4$, and $4$.

Using Vieta's formulas, the sum of the roots is:

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The product of the roots (considering that $4$ is a double root) gives us:

$$$$

Therefore, the values of $a$ and $b$ depend on $r$.

First, equate:

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

Multiplying through by the common denominator $(x + 1)(x^2 - 3)$ gives:

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

Now, by expanding and combining terms:

1. For $x^2$ coefficients: $1 = a + b$
2. For $x$ coefficients: $0 = b + c$
3. For constant terms: $-6 = -3a$

From the third equation, we have: $a = 2.$

Substituting back to find $b$ and $c$:

a. From $1 = a + b \Rightarrow b = -1$. b. From $0 = b + c \Rightarrow c = 1$.

Next, we solve the integral: $\int \left( \frac{2}{x + 1} - \frac{1x - 1}{x^2 - 3} \right) dx = 2 \ln |x + 1| - \int \frac{x - 1}{x^2 - 3} dx$.

The last integral can be solved by substitution, and combining all results gives the required answer.