Let $z = 2 + 3i$ and $w = 1 + i$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2001 - Paper 1

To find the modulus-argument form, we first calculate the modulus:

Next, we find the argument:

Therefore, in modulus-argument form:

Using De Moivre's Theorem:

Calculating:

and,

$$10 \cdot \frac{\pi}{3} = \frac{10\pi}{3} = 3\pi + \frac{\pi}{3} = \frac{\pi}{3} \quad \text{(as $3\pi$ is full rotations)}.

(1 + \sqrt{3}i)^{10} = 1024\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right).

= 1024\left(\frac{1}{2} + i \cdot \frac{\sqrt{3}}{2}\right) = 512 + 512\sqrt{3}i.

1. The inequality $|z + 1 - 2i| \leq 3$ represents a circle centered at $(-1, 2)$ with radius 3.
2. The inequalities for the argument describe a sector in the complex plane:
   • The line $heta = -\frac{\pi}{3}$ and $heta = \frac{\pi}{4}$.
3. Thus, the region where both conditions are satisfied is the intersection of the circle and the wedge defined by the two angles.

The equation can be rewritten as:

The fourth roots of $e^{i\pi}$ are given by:

Calculating:

• For $k = 0$: $z_0 = e^{i\frac{\pi}{4}}.$
• For $k = 1$: $z_1 = e^{i\frac{3\pi}{4}}.$
• For $k = 2$: $z_2 = e^{i\frac{5\pi}{4}}.$
• For $k = 3$: $z_3 = e^{i\frac{7\pi}{4}}.$

Thus, the solutions in modulus-argument form are:

Since triangle ABC is isosceles, the distances from points A ($z_1$) and C ($z_3$) to point B ($z_2$) must be equal. Therefore, it follows that:

Squaring both sides gives:

Since ABCD is a square, D is at a point obtained by rotating point C ($z_3$) by 90 degrees around point B ($z_2$). Thus, the complex representation for D is: