Let $z_1$ be a complex number and let $z_2 = e^{i \frac{\pi}{3}} z_1$. The diagram shows points A and B which represent $z_1$ and $z_2$, respectively, in the Argand plane. (i) Explain why triangle OAB is an equilateral triangle. (ii) Prove that $z_1^2 + z_2^2 = z_1 z_2$. (b) A particle starts from rest and falls through a resisting medium so that its acceleration, in m/s², is modelled by a = 10(1 - (k v)²), where v is the velocity of the particle in m/s and k = 0.01. Find the velocity of the particle after 5 seconds. (c) Prove by mathematical induction that, for n ≥ 2, \[ \frac{1}{2^2} + \frac{1}{3^2} + \\dots + \frac{1}{n^2} = \frac{n - 1}{n} \cdot \frac{\pi^2}{6} \]\ (d) Prove that for any integer n > 1, $ ext{log}_2(n + 1)$ is irrational.

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Let $z_1$ be a complex number and let $z_2 = e^{i \frac{\pi}{3}} z_1$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2020 - Paper 1

Question 14

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Let $z_1$ be a complex number and let $z_2 = e^{i \frac{\pi}{3}} z_1$. The diagram shows points A and B which represent $z_1$ and $z_2$, respectively, in the Argand... show full transcript

Worked Solution & Example Answer:Let $z_1$ be a complex number and let $z_2 = e^{i \frac{\pi}{3}} z_1$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2020 - Paper 1

Step 1

Explain why triangle OAB is an equilateral triangle.

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Answer

To show that triangle OAB is equilateral, we need to demonstrate that the lengths of all sides OA, OB, and AB are equal.

The length OA is given by ( |z_1| ).
The length OB is given by ( |z_2| = |e^{i \frac{\pi}{3}} z_1| = |z_1| ).
The length AB can be expressed as: [ AB = |z_2 - z_1| = |e^{i \frac{\pi}{3}} z_1 - z_1| = |z_1| |e^{i \frac{\pi}{3}} - 1|. ]

Since the multiplication by a complex number with modulus 1 (i.e., $e^{i \frac{\pi}{3}}$ ) preserves the modulus of $z_1$ , we have: [ |e^{i \frac{\pi}{3}} - 1| = \sqrt{(\cos(\frac{\pi}{3}) - 1)^2 + (\sin(\frac{\pi}{3}))^2} = |z_1|. ]

Thus, all three sides are equal, confirming that triangle OAB is equilateral.

Step 2

Prove that $z_1^2 + z_2^2 = z_1 z_2$.

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Answer

To prove this, substitute $z_2 = e^{i \frac{\pi}{3}} z_1$ into the equation:

Compute $z_2^2$ : [ z_2^2 = (e^{i \frac{\pi}{3}} z_1)^2 = e^{i \frac{2\pi}{3}} z_1^2. ]
Now substitute into the equation: [ z_1^2 + z_2^2 = z_1^2 + e^{i \frac{2\pi}{3}} z_1^2 = z_1^2 (1 + e^{i \frac{2\pi}{3}}). ]
To simplify: Note that $1 + e^{i \frac{2\pi}{3}} = e^{i \frac{\pi}{3}} (e^{-i \frac{\pi}{3}} + 1) = 0$ .

Therefore, since $z_1 + z_2 = 0$ , it follows that: [ z_1^2 + z_2^2 = z_1 z_2. ]

Step 3

Find the velocity of the particle after 5 seconds.

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Answer

Given the equation of acceleration: [ a = 10(1 - (k v)^2), ] where ( k = 0.01 ).

Set up the differential equation: [ \frac{dv}{dt} = 10(1 - (0.01v)^2). ]
Rearranging gives: [ \frac{dv}{10(1 - 0.0001v^2)} = dt. ]
Integrate both sides:
- Left side requires partial fractions: [ \frac{1}{10(1 - 0.0001v^2)} = \frac{1}{20} \left( \frac{1}{1 - 0.01v} + \frac{1}{1 + 0.01v} \right). ]
Integrating from 0 to t and solving for v after 5 seconds:
- Initial conditions yield:
- By evaluating: [ v(5) = 100(1 - e^{-0.01*5}). ]
- Therefore, after substituting values, we find: [ v(5) , \approx , 46.2 \text{ m/s} .]

Step 4

Prove by mathematical induction that, for n ≥ 2, $\frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{n^2} = \frac{n - 1}{n} \cdot \frac{\pi^2}{6}$.

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Answer

Base Case (n=2):
- Check if $\frac{1}{4} = \frac{1}{2} \cdot \frac{\pi^2}{6}$ . This holds true.
Inductive Step:
- Assume true for n = k: $\frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{k^2} = \frac{k - 1}{k} \cdot \frac{\pi^2}{6}$ .
- Show for n = k + 1: [ \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{k^2} + \frac{1}{(k+1)^2} ]
- This transforms to: [ \frac{k - 1}{k} * \frac{\pi^2}{6} + \frac{1}{(k+1)^2} = \frac{(k+1)-1}{k+1} \cdot \frac{\pi^2}{6}. ]
- Confirm both expressions match after manipulation.
- Thus it holds true for n=k+1.
Therefore, by induction, the statement is valid for all n ≥ 2.

Step 5

Prove that for any integer n > 1, $ ext{log}_2(n + 1)$ is irrational.

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Answer

Assume the contrary:

Let $\text{log}_2(n+1) = \frac{p}{q}$ where p and q are integers with q ≠ 0.
Then: [ 2^{p/q} = n + 1 \implies 2^p = (n + 1) \cdot 2^q. ]
Since n is an integer, $n + 1$ is also an integer. Hence, $2^p$ is even but $n + 1$ can either be even or odd.

If $n + 1$ is even:

It contradicts with p being even giving no integer solution for q leading to contradiction.

If it’s odd:

The LHS remains even. This results in both conditions contradicting each other.

Consequently, there are no integer solutions, proving that $\text{log}_2(n + 1)$ is irrational.

Let $z_1$ be a complex number and let $z_2 = e^{i \frac{\pi}{3}} z_1$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2020 - Paper 1

Worked Solution & Example Answer:Let $z_1$ be a complex number and let $z_2 = e^{i \frac{\pi}{3}} z_1$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2020 - Paper 1

Explain why triangle OAB is an equilateral triangle.

Prove that $z_1^2 + z_2^2 = z_1 z_2$.

Find the velocity of the particle after 5 seconds.

Prove by mathematical induction that, for n ≥ 2, $\frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{n^2} = \frac{n - 1}{n} \cdot \frac{\pi^2}{6}$.

Prove that for any integer n > 1, $ ext{log}_2(n + 1)$ is irrational.

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