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

First, calculate $z - 2$:

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

Next, we will divide by $w$:

$\frac{z - 2}{w} = \frac{3i}{1 - i}$

To simplify this expression, multiply the numerator and the denominator by the conjugate of the denominator:

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

Thus, we have:

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

Given that $p(x)$ has a zero at $r$ and a double zero at 4, we can express the polynomial as:

$p(x) = (x - r)(x - 4)^2$

Expanding this:

$p(x) = (x - r)(x^2 - 8x + 16) = x^3 - rx^2 - 8x^2 + 8rx + 16x - 16r$

Thus, equating coefficients you get:

$a = -r - 8, \, b = 8r + 16$

To find $r$, set $p(4)=0$:

$p(4) = (4 - r)(4 - 4)^2 = 0 \, \Rightarrow \text{this holds for any } r$

For the double zero, $p'(4) = 0$:

$p'(x) = 3x^2 + 2ax + b = 0$, Substituting back:

Solve for $a$, $b$ with given conditions.

To solve the integral, we first rewrite it in the form $\frac{a}{x + 1} + \frac{bx + c}{x^2 - 3}$:

1. We perform partial fraction decomposition:

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

2. Clearing denominators:

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

3. Expanding and equating coefficients:

Equating coefficients gives equations for $a$, $b$, and $c$. From the original fractions:

4. Finally, integrate the separate parts:

$\int \frac{a}{x + 1} \, dx + \int \frac{bx + c}{x^2 - 3} \, dx$