Let w be the complex number w = e^{2i rac{ ext{π}}{3}} - HSC - SSCE Mathematics Extension 2 - Question 16 - 2023 - Paper 1

Assuming triangle ABC is equilateral and oriented anticlockwise, vertices A, B, and C can be represented as:

• A = a,
• B = b,
• C = c.

For such a triangle in the complex plane, the following relationship holds:

$a + b + c = 0$

This can be expanded using w:

$b + c = -a$

Substituting gives us:

$a + b w + c w^2 = 0$

Therefore, the statement is verified.

For equilateral triangle ABC, the property holds that:

$(a - b)^2 + (b - c)^2 + (c - a)^2 = 0$

Expanding gives:

$a^2 + b^2 + c^2 - ab - ac - bc = 0$

Rearranging this leads to:

$a^2 + b^2 + c^2 = ab + ac + bc$

Thus, the equality is proven.

To show this inequality, we consider the function f(x) = x - ln(x). We find the derivative:

$f'(x) = 1 - \frac{1}{x}$

For x > 1, f'(x) > 0, hence f(x) is increasing. Evaluating at x = 1:

$f(1) = 1 - ln(1) = 1 > 0$

Thus, for x > 1, it follows that f(x) > 0 hence x > ln x. For 0 < x < 1, since ln(x) < 0, we can say that x > ln x still holds.

The proof starts with part (i), where we know:

$e^x > x$

This can be applied iteratively for integers. Using induction for n:

• Base case: For n = 1, the inequality holds.

Assume it holds for n, then for n + 1:

$e^{n+1} = e^n imes e > n! imes (n+1) > (n!)^{2}$

Thus, by induction, it holds for all positive integers n.

Given that the modulus of both w and z is 1, explain that:

1. The argument mentions that the division results in a quadrant.
2. The inequalities indicate that the argument is confined to the second quadrant of the complex plane.

The sketch would show the y-axis as a boundary, allowing for points where x < 0 and y > 0. The result reveals that:

$\frac{\pi}{2} < \text{Arg}(\frac{x + iy}{z}) < \pi$

Therefore, we can define the region accordingly.