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

To find $zw$, where $z = 3+i$ and $w = 2-5i$:

1. Calculate: $zw = (3+i)(2-5i) = 3\cdot2 + 3(-5i) + i\cdot2 + i(-5i)\ = 6 - 15i + 2i + 5 = 11 - 13i$

To find $\frac{w}{z}$:

1. Calculate: $\frac{w}{z} = \frac{2-5i}{3+i}$

2. Multiply the numerator and denominator by the conjugate of the denominator: $= \frac{(2-5i)(3-i)}{(3+i)(3-i)} = \frac{6 - 2i - 15i + 5}{9 + 1} = \frac{11 - 17i}{10} = 1.1 - 1.7i$

To express $\sqrt{3}-i$ in modulus-argument form:

1. Calculate the modulus: $r = |\sqrt{3}-i| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$

2. Calculate the argument: $\theta = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}$

Thus, the modulus-argument form is: $2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right)$

This step is identical to the previous one since we are expressing the same complex number:

The modulus-argument form is: $2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right)$

Using De Moivre's theorem:

$\left(\sqrt{3}-i\right)^7 = r^7\left(\cos(7\theta) + i\sin(7\theta)\right)$

1. Calculate $r^7 = 2^7 = 128$

2. Calculate the angle: $7\theta = 7(-\frac{\pi}{6}) = -\frac{7\pi}{6}$

Thus: $128\left(\cos\left(-\frac{7\pi}{6}\right) + i\sin\left(-\frac{7\pi}{6}\right)\right)$

3. Evaluating:
   • $\cos\left(-\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}$
   • $\sin\left(-\frac{7\pi}{6}\right) = -\frac{1}{2}$

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

To find all solutions to $z^3 = -1$:

1. Rewrite $-1$ in modulus-argument form:
   • Modulus: $1$
   • Argument: $\pi$

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

2. Using De Moivre's theorem: $z = r^{1/3}\left(\cos\left(\frac{\pi + 2k\pi}{3}\right) + i\sin\left(\frac{\pi + 2k\pi}{3}\right)\right) \, \text{for } k = 0, 1, 2$
   • Modulus: $r^{1/3} = 1^{1/3} = 1$
   • For $k=0$: $z_0 = \cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right) = \frac{1}{2} + i\frac{\sqrt{3}}{2}$
   • For $k=1$: $z_1 = \cos\left(\pi\right) + i\sin\left(\pi\right) = -1 + 0i$
   • For $k=2$: $z_2 = \cos\left(\frac{5\pi}{3}\right) + i\sin\left(\frac{5\pi}{3}\right) = \frac{1}{2} - i\frac{\sqrt{3}}{2}$

Thus, all solutions are:

1. $z_0 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$
2. $z_1 = -1$
3. $z_2 = \frac{1}{2} - i\frac{\sqrt{3}}{2}$

To find the center of the ellipse given by the equation: $|z-1-3i| + |z-9-3i| = 10$

The center is the midpoint between the foci:

• Foci: $(1, 3)$ and $(9, 3)$
• Center: $C = \left(\frac{1+9}{2}, 3\right) = (5, 3)$

Thus, the complex number corresponding to the center is: $z_C = 5 + 3i$

To sketch the ellipse described by: $|z-1-3i| + |z-9-3i| = 10$

1. The total sum is 10, indicating the length of the major axis is 10.
2. Distance between the foci: $d = 9 - 1 = 8$
3. Therefore, the semi-major axis $a = \frac{10}{2} = 5$
4. The semi-minor axis $b$ can be calculated using: $c = \sqrt{a^2-b^2} = 4$ (where $c$ is half the distance between the foci) Hence, $b = \sqrt{a^2 - c^2} = \sqrt{25 - 16} = 3$
5. Thus, the lengths of the axes are:
   • Major axis length: $10$
   • Minor axis length: $6$

To find the range of $\arg(z)$ for the ellipse:

1. The foci of the ellipse are at $1 + 3i$ and $9 + 3i$.
2. The ellipse will have a minimum and maximum argument depending on these foci:
   • At $z=1+3i$, $\arg(z) = \frac{\pi}{2}$
   • At $z=9+3i$, $\arg(z) = 0$

Thus, the range of the argument corresponds to: $\left(0, \frac{\pi}{2}\right)$