Let $z=3+i$ and $w=2-5i$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2006 - Paper 1

Next, calculate $zw$:

$zw = (3 + i)(2 - 5i) = 3 \cdot 2 + 3 \cdot (-5i) + i \cdot 2 + i \cdot (-5i)$

This simplifies to:

$6 - 15i + 2i + 5 = 11 - 13i$

So, $zw = 11 - 13i$.

To find $\frac{w}{z}$, perform the division:

$\frac{w}{z} = \frac{2 - 5i}{3 + i}$

Multiply the numerator and denominator by the complex conjugate of the denominator:

$= \frac{(2 - 5i)(3 - i)}{(3 + i)(3 - i)} = \frac{6 - 2i - 15i + 5}{9 + 1} = \frac{11 - 17i}{10} = 1.1 - 1.7i$

Thus, $\frac{w}{z} = 1.1 - 1.7i$.

First, calculate the modulus:

$r = |\sqrt{3}-i| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$

Next, calculate the argument:

$\theta = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}$

Thus, in modulus-argument form, $\sqrt{3}-i = 2 \text{cis}\left(-\frac{\pi}{6}\right)$.

This step is identical to part (i). Hence,

$\left(\sqrt{3}-i\right) = 2 \text{cis}\left(-\frac{\pi}{6}\right)$.

Now we use De Moivre's theorem:

$\left(\sqrt{3}-i\right)^7 = r^7 \text{cis}(7\theta) = 2^7 \text{cis}\left(7 \cdot -\frac{\pi}{6}\right) = 128 \text{cis}\left(-\frac{7\pi}{6}\right)$.

Next, express in $x + iy$:

$\text{cis}\left(-\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2} - \frac{1}{2}i$

Therefore:

$\left(\sqrt{3}-i\right)^7 = 128\left(-\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) = -64\sqrt{3} - 64i$.

To find the solutions to $z^3 = -1$, rewrite this in modulus-argument form:

$-1 = 1\text{cis} \pi$.

Finding the cube roots involves:

$z_k = \sqrt[3]{1}\text{cis}\left(\frac{\pi + 2k\pi}{3}\right), \quad k = 0, 1, 2$.

Calculating the roots:

1. For $k=0$: $z_0 = 1\text{cis}\left(\frac{\pi}{3}\right)$
2. For $k=1$: $z_1 = 1\text{cis}\left(\pi\right)\, = -1$
3. For $k=2$: $z_2 = 1\text{cis}\left(\frac{5\pi}{3}\right)$

Thus, the solutions are:

$z_0 = 1\text{cis}\left(\frac{\pi}{3}\right), z_1 = -1, z_2 = 1\text{cis}\left(\frac{5\pi}{3}\right).$

The center of the ellipse corresponds to the average of the foci:

$\text{Centre} = \frac{(1 + 9) + (3i + 3i)}{2} = \frac{10}{2} + 3i = 5 + 3i.$

The equation represents an ellipse centered at $5 + 3i$ with foci at $1 + 3i$ and $9 + 3i$. The distance between the foci is 8.

The total distance of $10$ provides the lengths of the axes:

• Major axis length: $10$
• Minor axis length: $\sqrt{10^2 - 8^2} = \sqrt{36} = 6$.

Draw the ellipse based on the center and axes provided.

To find the range of $\arg(z)$, consider the angle subtended by the endpoints of the line segment along the major axis:

At focus $1 + 3i$: $\arg(1 + 3i) = \tan^{-1}(3)$

At focus $9 + 3i$: $\arg(9 + 3i) = \tan^{-1}(\frac{3}{9}) = \tan^{-1}(\frac{1}{3})$

Thus, the range of $\arg(z)$ corresponds to:

$\left[\tan^{-1}(\frac{1}{3}), \tan^{-1}(3)\right].$