Let $w$ be the complex number $w = e^{\frac{2\pi i}{3}}$. (i) Show that $1 + w + w^2 = 0$. The vertices of a triangle can be labelled $A$, $B$ and $C$ in anticlockwise or clockwise direction, as shown. (ii) Show that if triangle $ABC$ is anticlockwise and equilateral, then $a + bv + cw = 0$. (iii) It can be shown that if triangle $ABC$ is clockwise and equilateral, then $a + bv + cw = 0$. (Do NOT prove this.) Show that if $ABC$ is an equilateral triangle, then $a^2 + b^2 + c^2 = ab + bc + ca$. (b) Prove that $x > 1$ implies $x > \ln x$, for $x > 0$. (ii) Using part (i), or otherwise, prove that for all positive integers $n$, e^{nx} > (n!)^2.$ (c) The complex numbers $w$ and $z$ both have modulus 1, and $\frac{\pi}{2} < \text{Arg}(\frac{z}{w}) < \pi$, where Arg denotes the principal argument. For real numbers $x$ and $y$, consider the complex number $\frac{x + iy}{z}$. On an $xy$-plane, clearly sketch the region that contains all points $(x, y)$ for which $\frac{\pi}{2} < \text{Arg}(\frac{x + iy}{z}) < \pi$.

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Let $w$ be the complex number $w = e^{\frac{2\pi i}{3}}$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2023 - Paper 1

Question 16

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Let $w$ be the complex number $w = e^{\frac{2\pi i}{3}}$. (i) Show that $1 + w + w^2 = 0$. The vertices of a triangle can be labelled $A$, $B$ and $C$ in anticlock... show full transcript

Worked Solution & Example Answer:Let $w$ be the complex number $w = e^{\frac{2\pi i}{3}}$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2023 - Paper 1

Step 1

Show that $1 + w + w^2 = 0$

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Answer

To show that $1 + w + w^2 = 0$ , we know that $w = e^{\frac{2\pi i}{3}}$ .

Calculating $w^2$ : $w^2 = (e^{\frac{2\pi i}{3}})^2 = e^{\frac{4\pi i}{3}}.$

Now we need to find the sum: $1 + w + w^2 = 1 + e^{\frac{2\pi i}{3}} + e^{\frac{4\pi i}{3}}.$

Using properties of complex exponentials: $1 + e^{\frac{2\pi i}{3}} + e^{\frac{4\pi i}{3}} = 0$ This is because these points represent vertices of an equilateral triangle in the complex plane, summing to zero.

Step 2

Show that if triangle $ABC$ is anticlockwise and equilateral, then $a + bv + cw = 0$

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Answer

For triangle $ABC$ to be equilateral and anticlockwise, we can rotate the points in the complex plane such that:

The point $A$ can be taken as $a$ .
The point $B$ can be defined as $b = e^{\frac{2\pi i}{3}}a$ (90 degree rotation).
The point $C$ can be defined as $c = e^{\frac{4\pi i}{3}}a$ . Thus, we have:

a + bv + cw &= a + e^{\frac{2\pi i}{3}} a + e^{\frac{4\pi i}{3}} a \ &= a(1 + e^{\frac{2\pi i}{3}} + e^{\frac{4\pi i}{3}}) \ &= 0 \quad \text{(as shown in part (i))}. \end{align*}$$

Step 3

Show that if $ABC$ is an equilateral triangle, then $a^2 + b^2 + c^2 = ab + bc + ca$

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Answer

For an equilateral triangle, using the property of symmetric sums, we can say:

a^2 + b^2 + c^2 = ab + bc + ca \ \Rightarrow (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \ \Rightarrow 0 = 0 \text{ if } a = b = c. \end{align*}$$ This further proves that this relationship holds true for any equilateral triangle.

Step 4

Prove that $x > 1$ implies $x > \ln x$, for $x > 0$

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Answer

Define the function: $f(x) = x - \ln x.$

Calculating its derivative gives: $f'(x) = 1 - \frac{1}{x}.$

For $x > 1$ , we have $f'(x) > 0$ , indicating that the function is increasing. Also, at $x = 1$ , we find: $f(1) = 1 - \ln(1) = 1 > 0.$ Therefore, for all $x > 1, f(x) > 0 \Rightarrow x > \ln x.$

Step 5

Prove that for all positive integers $n$, $e^{nx} > (n!)^2$

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Answer

Using induction, we let the base case for $n = 1$ be true: $e^x > 1 > 1^2.$

Assuming true for $n$ , we can show for $n + 1$ : $e^{(n + 1)x} = e^x imes e^{nx} > 1 \times (n!)^2 = (n!)^2.$

Hence, by inductive reasoning, it holds for all positive integers.

Step 6

Sketch the region that contains all points $(x, y)$ where $\frac{\pi}{2} < \text{Arg}(\frac{x + iy}{z}) < \pi$

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Answer

The argument of complex number $rac{x + iy}{z}$ is a representation of the angle in polar coordinates. We want to identify the region between angles:

$\frac{\pi}{2}$ represents the positive imaginary axis.
$\pi$ represents the negative real axis. Thus, the required region is the left half of the complex plane above the real axis, excluding the line $\text{Arg} = \frac{\pi}{2}$ and $\text{Arg} = \pi$ .

Let $w$ be the complex number $w = e^{\frac{2\pi i}{3}}$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2023 - Paper 1

Worked Solution & Example Answer:Let $w$ be the complex number $w = e^{\frac{2\pi i}{3}}$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2023 - Paper 1

Show that $1 + w + w^2 = 0$

Show that if triangle $ABC$ is anticlockwise and equilateral, then $a + bv + cw = 0$

Show that if $ABC$ is an equilateral triangle, then $a^2 + b^2 + c^2 = ab + bc + ca$

Prove that $x > 1$ implies $x > \ln x$, for $x > 0$

Prove that for all positive integers $n$, $e^{nx} > (n!)^2$

Sketch the region that contains all points $(x, y)$ where $\frac{\pi}{2} < \text{Arg}(\frac{x + iy}{z}) < \pi$

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