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

Assume triangle $ABC$ is anticlockwise and equilateral. The vertices can be expressed in terms of complex numbers $A$, $B$, and $C$ as follows. Using the properties of equilateral triangles in the complex plane, we find that:

$a + b + c = 0$
This implies
a + bv + cw = 0,
by the relationship of the vertices in the complex plane.

For triangle $ABC$ being equilateral, we apply the identity:

$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$

Since $a + b + c = 0$, it follows that:

$0 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
Rearranging gives us:

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

To prove this, consider the function:

$f(x) = x - ext{ln}(x)$
Taking the derivative, we find:

f'(x) = 1 - rac{1}{x} This shows that $f'(x) > 0$ for $x > 1$. Thus, the function is increasing, and since $f(1) = 0 > 0$, it follows that:

$x > ext{ln}(x) ext{ for } x > 0.$

Using induction, assume for $k ext{ positive integer, } e^k > k!$.
For $n = k + 1$:

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

To determine the region where:

$\frac{ ext{π}}{2} < ext{Arg} \left( \frac{x + ext{i}y}{z} \right) < ext{π}$

We have to take into account the angle properties in the complex plane. The inequalities describe a sector where:

1. The real part must be negative, excluded in the first quadrant.
2. The second quadrant will satisfy valid $y < 0$ values. Thus, the sketch will show this bounded area.