Let $z = oot{3} - i.$ (i) Express $z$ in modulus-argument form - HSC - SSCE Mathematics Extension 2 - Question 11 - 2016 - Paper 1

To show that $z^6$ is real, we use the formula for powers in modulus-argument form:

$z^n = |z|^n \left(\cos(n \cdot \text{Arg}(z)) + i \sin(n \cdot \text{Arg}(z))\right).$

For $n=6$:

$z^6 = 2^6 \left(\cos\left(6 \cdot -\frac{\pi}{6}\right) + i \sin\left(6 \cdot -\frac{\pi}{6}\right)\right) = 64\left(\cos(-\pi) + i \sin(-\pi)\right) = 64(-1 + 0i) = -64,$ which is real.

For $z^n$ to be purely imaginary, the cosine term must equal zero. This occurs when:

$n \cdot -\frac{\pi}{6} = \frac{\pi}{2} + k\pi, \text{ for any integer } k.$

Solving:

$n = -3 + 6k.$

For a positive integer, let $k = 1$:
$n = -3 + 6(1) = 3.$

Thus, $n=3$ is a solution.

To find (\int x e^{-2x} , dx,) we can use integration by parts. Let:

• $u = x$, so $du = dx$
• $dv = e^{-2x} dx$, so $v = -\frac{1}{2} e^{-2x}.$

Using the integration by parts formula:

$\int u \, dv = uv - \int v \, du,$

we have:

$\int x e^{-2x} \, dx = -\frac{1}{2} x e^{-2x} - \int -\frac{1}{2} e^{-2x} \, dx$

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

To find (\frac{dy}{dx}), we differentiate implicitly:

$\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(2xy).$

This gives:

$3x^2 + 3y^2 \frac{dy}{dx} = 2\left( y + x \frac{dy}{dx} \right).$

Rearranging gives:

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

$\frac{dy}{dx}(3y^2 - 2x) = 2y - 3x^2,$

thus:

$\frac{dy}{dx} = \frac{2y - 3x^2}{3y^2 - 2x}.$