Consider the complex numbers $w = -1 + 4i$ and $z = 2 - i$ - HSC - SSCE Mathematics Extension 2 - Question 11 - 2020 - Paper 1

To evaluate the product of the complex numbers $w$ and $z$, we calculate:

$wz = (-1 + 4i)(2 - i).$ Using the distributive property (FOIL), we get:

$wz = -2 + i + 8i - 4i^2.$ Since $i^2 = -1$, we have:

$-4i^2 = 4$ Thus,

$wz = -2 + 9i + 4 = 2 + 9i.$

For integration by parts, we choose:

• Let $u = \ln x$ ⇒ $du = \frac{1}{x} \, dx$
• Let $dv = x \, dx$ ⇒ $v = \frac{x^2}{2}$

Now, applying the integration by parts formula: $\int u \, dv = uv - \int v \, du,$
we get:

$\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx.$
This simplifies to:

$\int x \ln x \, dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C.$

Evaluating from 1 to $e$:

$\left[ \frac{e^2}{2} \ln e - \frac{e^2}{4} \right] - \left[ \frac{1^2}{2} \ln 1 - \frac{1^2}{4} \right] = \left[ \frac{e^2}{2} - \frac{e^2}{4} \right] - \left[ 0 - \frac{1}{4} \right].$

Thus,

$\frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4}.$

Given the acceleration: $a = y^2 + v,$
we know that acceleration is the derivative of velocity: $a = \frac{dv}{dx}.$
Thus,

$\frac{dv}{dx} = y^2 + v.$
Rearranging gives: $\frac{dx}{dv} = \frac{1}{y^2 + v}.$
To find $x$, integrate with respect to $v$:

$x = \int \frac{1}{v + y^2} \, dv + C.$
Integrating yields: $x = \ln(v + y^2) + C.$ Considering that at $x = 0$, $v = 1$ gives: $0 = \ln(1 + y^2) + C \implies C = -\ln(1 + y^2).$
So,

$x = \ln(v + 1) - \ln(1 + y^2) = \ln\left(\frac{v + 1}{1 + y^2}\right).$

To determine when the vectors $u$ and $v$ are perpendicular, we compute:

$u + v = (-2i - j + 3k) + (\rho i + j + 2k) = (-2 + \rho)i + (0)j + (3 + 2)k.$
For perpendicular vectors, their dot product must equal zero:

$u \cdot v = (-2 + \rho)(0) + (0)(1) + (5)(3) = 5(2 + \rho) = 0.$ Thus, $\rho = -2.$

Therefore, the value of $\rho$ that makes $u + v$ perpendicular is: $\rho = -2.$

The equation takes the form: $z^2 + 3z + (3 - i) = 0.$
Using the quadratic formula: $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ here, $a = 1$, $b = 3$, and $c = 3 - i$. Thus,

$\Delta = 9 - 4(1)(3 - i) = 9 - 12 + 4i = -3 + 4i.$
Now, find the square root: $\sqrt{-3 + 4i} = 1 + 2i.$ Substituting back into the quadratic formula: $z = \frac{-3 \pm (1 + 2i)}{2}.$ Calculating yields: $z_1 = \frac{-3 + 1 + 2i}{2} = -1 + i, \\ z_2 = \frac{-3 - 1 - 2i}{2} = -2 - i.$
Thus, the final answers are: $z = -1 + i \, \text{and} \, z = -2 - i.$