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

To find $zw$, multiply $z = 3 + i$ with $w = 2 - 5i$:

$zw = (3+i)(2-5i) = 6 - 15i + 2i + 5 = 11 - 13i$

To find $\frac{w}{z}$, you divide $w$ by $z$:

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

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

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

First, calculate the modulus:

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

Next, calculate the argument:

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

Thus, the modulus-argument form is:

$\sqrt{3} - i = 2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right)$

This is already done in part (i), but restating:

$\sqrt{3} - i = 2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right)$

Using De Moivre's Theorem:

$\left(\sqrt{3}-i\right)^7 = 2^7\left(\cos\left(-\frac{7\pi}{6}\right) + i\sin\left(-\frac{7\pi}{6}\right)\right) = 128\left(\cos\left(-\frac{7\pi}{6}\right) + i\sin\left(-\frac{7\pi}{6}\right)\right)$

Now, calculate:

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

Rewriting $-1$ in modulus-argument form:

$-1 = 1\left(\cos\pi + i\sin\pi\right)$

Using De Moivre's Theorem, we find the cube roots:

1. The principal root:
   • $z_1 = 1\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = \frac{1}{2} + \frac{\sqrt{3}}{2}i$
2. The other roots are:
   • $z_2 = 1\left(\cos\left(\frac{\pi + 2\pi}{3}\right) + i\sin\left(\frac{\pi + 2\pi}{3}\right)\right)$ and derive these further.

The equation can be rewritten as:

$|z - (5 - 3i)| = 10$

The center of the ellipse is at the point $(5, 3)$ or in complex form: $5 + 3i.$

In the equation given, the distance from the center to the focus is denoted as $a = 10$ and the distance between the foci is $2b$ where $b$ can be derived from the distances:

Thus, the major axis length is $20$ (for $x$) and the minor axis length can be derived from the standard ellipse equation.

Simply speaking, the sketch will portray an ellipse centered at $(5, 3)$.

The range of $\arg(z)$ will span from the angle corresponding to the leftmost point to the angle corresponding to the rightmost point.

Thus: $[\tan^{-1}(\frac{3-b}{5-a}), \tan^{-1}(\frac{3+b}{5+a})]$ where $(a, b)$ represents the ellipse dimensions.