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

To express $\frac{z - 2}{w}$ in the form $x + iy$, we first calculate $z - 2$:

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

Now substituting $w = 1 - i$, we have:

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

Next, multiply the numerator and the denominator by the conjugate of the denominator:

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

Thus, simplifying gives:

$= -\frac{3}{2} + \frac{3}{2}i$

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

Given that $p(x) = x^3 + ax^2 + b$ has a zero at $r$ and a double zero at 4, we can utilize the fact that the polynomial can be expressed as:

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

Expanding $(x - 4)^2$ gives:

$(x - 4)(x - 4) = x^2 - 8x + 16$

Thus,

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

Setting the coefficients equal:

• For $x^2$: $a = -(r + 8)$
• For $x^1$: $b = 16 + 8r$
• For the constant: $-16r = 0 \Rightarrow r = 0$

Substituting r back, we find:

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

Therefore, the values are $a = -8$, $b = 16$, and $r = 0$.

To find the integral, we first rewrite:

$\frac{x^2 - 6}{(x + 1)(x^2 - 3)}$

Using partial fraction decomposition:

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

Multiplying through by the denominator cancels:

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

Expanding gives:

$x^2 - 6 = ax^2 - 3a + bx^2 + cx + b$

By gathering like terms:

$(a + b)x^2 + cx + (-3a + b) = x^2 - 6$

Equating coefficients yields:

1. $a + b = 1$
2. $c = 0$
3. $-3a + b = -6$

Solving gives:

• From $a + b = 1,$ substituting $b = 1 - a$ in the third equation: $-3a + (1 - a) = -6 \Rightarrow -4a + 1 = -6 \Rightarrow -4a = -7 \Rightarrow a = \frac{7}{4}, \, b = -\frac{3}{4}, \, c = 0$

Finally, substituting back, we have:$\int (\frac{7/4}{x+1} - \frac{3/4x}{x^2-3}) dx$,

Integrating yields:

$= \frac{7}{4} \ln|x + 1| - \frac{3}{8} \ln|x^2 - 3| + C$