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david.wang28

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Re: 4U Maths Question Thread
« Reply #2115 on: November 21, 2018, 04:53:39 pm »
0
Hello,
Can anyone please give me a detailed answer for 4 d) and 4 f) ASAP? Thanks 😁
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2116 on: November 21, 2018, 05:38:14 pm »
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Hello,
Can anyone please give me a detailed answer for 4 d) and 4 f) ASAP? Thanks 😁
I don't see any attachment?

david.wang28

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Re: 4U Maths Question Thread
« Reply #2117 on: November 21, 2018, 08:26:21 pm »
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I don't see any attachment?
Sorry, my bad. Here's the attachment of the questions
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2118 on: November 21, 2018, 08:31:34 pm »
+3

\begin{align*} \overline{5z} &= \overline{5(x+iy)}\\ &= \overline{5x+5iy}\\ &= 5x-5iy\\ &= 5(x-iy)\\ &= 5\overline{z} \end{align*}
_________________________________________________
\begin{align*} \overline{\left(\frac{1}{z}\right)}&= \overline{\left( \frac{1}{x+iy} \right)}\\ &= \overline{\frac{x-iy}{x^2+y^2}}\\ &= \frac{x+iy}{x^2+y^2}\\ &= \frac{1}{x-iy}\\ &= \frac{1}{\overline{z}} \end{align*}
The steps used are only common techniques in the MX2 course. If you require a step to be expanded on, please state that step.
« Last Edit: November 21, 2018, 08:34:42 pm by RuiAce »

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Re: 4U Maths Question Thread
« Reply #2119 on: November 21, 2018, 08:39:11 pm »
0

\begin{align*} \overline{5z} &= \overline{5(x+iy)}\\ &= \overline{5x+5iy}\\ &= 5x-5iy\\ &= 5(x-iy)\\ &= 5\overline{z} \end{align*}
_________________________________________________
\begin{align*} \overline{\left(\frac{1}{z}\right)}&= \overline{\left( \frac{1}{x+iy} \right)}\\ &= \overline{\frac{x-iy}{x^2+y^2}}\\ &= \frac{x+iy}{x^2+y^2}\\ &= \frac{1}{x-iy}\\ &= \frac{1}{\overline{z}} \end{align*}
The steps used are only common techniques in the MX2 course. If you require a step to be expanded on, please state that step.
Thanks for the solutions :)
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david.wang28

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Re: 4U Maths Question Thread
« Reply #2120 on: November 22, 2018, 08:24:57 pm »
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Hello,
Can anyone please give me a detailed answer for 9 d) and 9 c) ASAP? Thanks 😁
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2121 on: November 22, 2018, 08:41:41 pm »
+6
Hello,
Can anyone please give me a detailed answer for 9 d) and 9 c) ASAP? Thanks 😁
I will only do the first one mentioned as they both follow the exact same procedure.
\[ \text{Note that }(1+i)^2 = 1+2i+i^2 = 2i. \]





\begin{align*} x &= \frac{2+2i \pm (4 - 4i)}{2i}\\ &= \frac{6-2i}{2i}, \frac{2-6i}{2i}\\ &=3+i, -1-3i\end{align*}

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Re: 4U Maths Question Thread
« Reply #2122 on: November 24, 2018, 06:54:33 pm »
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Heyyy. Currently stuck on a question and thought I'd ask some geniuses. Also does one even know the first step to take in questions like these?
haha thanks heaps guys!

If:
Prove that:
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2123 on: November 24, 2018, 07:38:19 pm »
+3
Heyyy. Currently stuck on a question and thought I'd ask some geniuses. Also does one even know the first step to take in questions like these?
haha thanks heaps guys!

If:
Prove that:
This particular question is more of a trick. It relies on whether you've seen ratios of lengths before. If you haven't, then the intuition to use a geometric approach becomes far less obvious.
Because the ratios of lengths are 3-4-5, this is hinting at some kind of Pythagoras' Theorem result.


Whilst it turns out it doesn't look like that for this question, the important thing to recall is that \(z_2-z_1\) is just the vector from \(z_1\) to \(z_2\), and hence its length \(|z_2-z_1|\) just represents the remaining side in the triangle, whose vertices are \(0\), \(z_1\) and \(z_2\).





Note that the cases of plus/minus didn't matter as it gets made positive anyway after squaring.

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Re: 4U Maths Question Thread
« Reply #2124 on: November 24, 2018, 08:20:04 pm »
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That is amazing! Wow. Thank you heaps. I'll be on the look out for Pythagorean Triads next time haha. Thank you again!!!
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david.wang28

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Re: 4U Maths Question Thread
« Reply #2125 on: December 02, 2018, 05:22:00 pm »
0
Hello,
Can anyone please give me a detailed answer for Q4 to Q9 ASAP (I have included some of my working out, Q8 and Q9 I cannot do at all)?. Thanks :)
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Re: 4U Maths Question Thread
« Reply #2126 on: December 03, 2018, 11:23:33 pm »
+6
Hello,
Can anyone please give me a detailed answer for Q4 to Q9 ASAP (I have included some of my working out, Q8 and Q9 I cannot do at all)?. Thanks :)
Hey, don't usually go through the MX2 thread as I didn't do MX2 but considering no one's answered it, I'll give them a crack!

I'll provide hints for you and then a full solution to some of the questions in the spoiler! :)

Q4
Hint 1: If you want to prove that a triangle is right angled, then you may like to use Pythagoras' Theorem! Perhaps, try showing that:
\[ \left|OQ\right|^2 = \left|OP\right|^2 + \left|PQ\right|^2.\]

Q4 - full solution
As hinted, we'll try to employ Pythagoras' Theorem.
First note that given: \(z = x + iy\), we have \(\left|OP\right| = \sqrt{x^2 + y^2}\) and:
\[ \boxed{\left|OP\right|^2 = x^2 + y^2}\]

We also have:
\[\begin{align*}z + iz &= (x + iy) + i(x + iy) \\ &= (x - y) + i(x + y).\end{align*}\]
So we get:
\[ \boxed{\left|OQ\right|^2 = (x - y)^2 + (x + y)^2}\]

Finally, the line segment \(PQ\) is just \(Q - P = ((x - y) - x) + i((x + y) - y) = -y + ix.\)

So we get:
\[ \boxed{\left|PQ\right|^2 = y^2 + x^2}\]

Now, we know that:
\[ \begin{align*}\left|OQ\right|^2 &= (x - y)^2 + (x + y)^2 \\ &= x^2 - 2xy + y^2 + x^2 + 2xy + y^2 \\ &= (x^2 + y^2) + (x^2 + y^2) \\ &= \left|OP\right|^2 + \left|PQ\right|^2.\end{align*}\]

Q5
Q5 - full solution
Let:
\[z_1 = r\text{cis}\theta\] and \[z_2 = r\text{cis}\left(\theta + \frac{\pi}{3}\right)\]

We then have the following (using De Moivre's Theorem):
\[ z_1^2 = r^2\text{cis}(2\theta), \quad z_2^2 = r^2\text{cis}\left(2\theta + \frac{2\pi}{3}\right)\]
Adding these two gives us:
\[ \begin{align*}z_1^2 + z_2^2 &= r^2\text{cis}(2\theta) + r^2\text{cis}\left(2\theta + \frac{2\pi}{3}\right) \\ &= r^2\left(\text{cis}(2\theta) + \text{cis}\left(2\theta + \frac{2\pi}{3}\right)\right) \\ &= r^2\text{cis}\left(2\theta + \frac{\pi}{3}\right) \\ &= r^2\text{cis}\left(\theta + \left(\theta + \frac{\pi}{3}\right)\right) \\ &= r\text{cis}\theta \times r\text{cis}\left(\theta + \frac{\pi}{3}\right) \\ &= z_1z_2\end{align*}\]

Q6
Hint 1: You may like to take a geometric approach :) Try drawing up \(z_1\) on an Argand diagram and from the nose of \(z_1\), draw \(z_2\). The result gives us: \(z_1 + z_2\).
Hint 2: The triangle inequality (see Q8) states that:
\[ \left|x+y\right| \leq \left|x\right| + \left|y\right|\]
Q6 - full solution
We note that:
\[ \left|z_1\right| = \left|(z_1 - z_2) + z_2\right|\]
And by the triangle inequality, we see that:
\[ \left|z_1\right| \leq \left|z_1 - z_2\right| + \left|z_2\right|\]

By a similar argument, the same can be said for \(z_2\). That is:
\[ \left|z_2\right| \leq \left|z_1 - z_2\right| + \left|z_1\right|\]

We see that:
\[ \left|z_1\right| - \left|z_2\right| \leq \left|z_1 - z_2\right|, \quad \left|z_2\right| - \left|z_1\right| \leq \left|z_1 - z_2\right|\]
which implies that:
\[ \left|\left|z_1\right|-\left|z_2\right|\right| \leq \left|z_1 - z_2\right|.\]

Q8
Hint: Best bet would be to prove it by induction.
Q8 - full solution
Base case is just the simple triangle inequality.

When \(n = k\), we have:
\[ \left|z_1 + z_2 + z_3 + \dots + z_k\right| \leq \left|z_1\right| + \left|z_2\right| + \left|z_3\right| + \dots + \left|z_k\right|.\]

When \(n = k+1\), we have:
\[ \begin{align*}\left|z_1 + z_2 + z_3 + \dots + z_k + z_{k+1}\right| &\leq \left|z_1 + z_2 + z_3 + \dots + z_k\right| + \left|z_{k+1}\right| \tag{by our triangle inequality} \\ &\leq \left|z_1\right| + \left|z_2\right| + \left|z_3\right| + \dots + \left|z_{k+1}\right|\end{align*}\]
which completes the proof.

Q9
Q9 - full solution
If the triangles \(OBA\) and \(OID\) are similar, then it follows that:
\[ \frac{OD}{OA} = \frac{OI}{OB}\]

and thus, we have:
\[ \frac{OD}{OI} = \frac{OA}{OB} = \frac{z_1}{z_2} = OD.\]

david.wang28

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Re: 4U Maths Question Thread
« Reply #2127 on: December 04, 2018, 08:23:09 pm »
0
Hey, don't usually go through the MX2 thread as I didn't do MX2 but considering no one's answered it, I'll give them a crack!

I'll provide hints for you and then a full solution to some of the questions in the spoiler! :)

Q4
Hint 1: If you want to prove that a triangle is right angled, then you may like to use Pythagoras' Theorem! Perhaps, try showing that:
\[ \left|OQ\right|^2 = \left|OP\right|^2 + \left|PQ\right|^2.\] Thank you for the help kind sir!

Q4 - full solution
As hinted, we'll try to employ Pythagoras' Theorem.
First note that given: \(z = x + iy\), we have \(\left|OP\right| = \sqrt{x^2 + y^2}\) and:
\[ \boxed{\left|OP\right|^2 = x^2 + y^2}\]

We also have:
\[\begin{align*}z + iz &= (x + iy) + i(x + iy) \\ &= (x - y) + i(x + y).\end{align*}\]
So we get:
\[ \boxed{\left|OQ\right|^2 = (x - y)^2 + (x + y)^2}\]

Finally, the line segment \(PQ\) is just \(Q - P = ((x - y) - x) + i((x + y) - y) = -y + ix.\)

So we get:
\[ \boxed{\left|PQ\right|^2 = y^2 + x^2}\]

Now, we know that:
\[ \begin{align*}\left|OQ\right|^2 &= (x - y)^2 + (x + y)^2 \\ &= x^2 - 2xy + y^2 + x^2 + 2xy + y^2 \\ &= (x^2 + y^2) + (x^2 + y^2) \\ &= \left|OP\right|^2 + \left|PQ\right|^2.\end{align*}\]

Q5
Q5 - full solution
Let:
\[z_1 = r\text{cis}\theta\] and \[z_2 = r\text{cis}\left(\theta + \frac{\pi}{3}\right)\]

We then have the following (using De Moivre's Theorem):
\[ z_1^2 = r^2\text{cis}(2\theta), \quad z_2^2 = r^2\text{cis}\left(2\theta + \frac{2\pi}{3}\right)\]
Adding these two gives us:
\[ \begin{align*}z_1^2 + z_2^2 &= r^2\text{cis}(2\theta) + r^2\text{cis}\left(2\theta + \frac{2\pi}{3}\right) \\ &= r^2\left(\text{cis}(2\theta) + \text{cis}\left(2\theta + \frac{2\pi}{3}\right)\right) \\ &= r^2\text{cis}\left(2\theta + \frac{\pi}{3}\right) \\ &= r^2\text{cis}\left(\theta + \left(\theta + \frac{\pi}{3}\right)\right) \\ &= r\text{cis}\theta \times r\text{cis}\left(\theta + \frac{\pi}{3}\right) \\ &= z_1z_2\end{align*}\]

Q6
Hint 1: You may like to take a geometric approach :) Try drawing up \(z_1\) on an Argand diagram and from the nose of \(z_1\), draw \(z_2\). The result gives us: \(z_1 + z_2\).
Hint 2: The triangle inequality (see Q8) states that:
\[ \left|x+y\right| \leq \left|x\right| + \left|y\right|\]
Q6 - full solution
We note that:
\[ \left|z_1\right| = \left|(z_1 - z_2) + z_2\right|\]
And by the triangle inequality, we see that:
\[ \left|z_1\right| \leq \left|z_1 - z_2\right| + \left|z_2\right|\]

By a similar argument, the same can be said for \(z_2\). That is:
\[ \left|z_2\right| \leq \left|z_1 - z_2\right| + \left|z_1\right|\]

We see that:
\[ \left|z_1\right| - \left|z_2\right| \leq \left|z_1 - z_2\right|, \quad \left|z_2\right| - \left|z_1\right| \leq \left|z_1 - z_2\right|\]
which implies that:
\[ \left|\left|z_1\right|-\left|z_2\right|\right| \leq \left|z_1 - z_2\right|.\]

Q8
Hint: Best bet would be to prove it by induction.
Q8 - full solution
Base case is just the simple triangle inequality.

When \(n = k\), we have:
\[ \left|z_1 + z_2 + z_3 + \dots + z_k\right| \leq \left|z_1\right| + \left|z_2\right| + \left|z_3\right| + \dots + \left|z_k\right|.\]

When \(n = k+1\), we have:
\[ \begin{align*}\left|z_1 + z_2 + z_3 + \dots + z_k + z_{k+1}\right| &\leq \left|z_1 + z_2 + z_3 + \dots + z_k\right| + \left|z_{k+1}\right| \tag{by our triangle inequality} \\ &\leq \left|z_1\right| + \left|z_2\right| + \left|z_3\right| + \dots + \left|z_{k+1}\right|\end{align*}\]
which completes the proof.

Q9
Q9 - full solution
If the triangles \(OBA\) and \(OID\) are similar, then it follows that:
\[ \frac{OD}{OA} = \frac{OI}{OB}\]

and thus, we have:
\[ \frac{OD}{OI} = \frac{OA}{OB} = \frac{z_1}{z_2} = OD.\]
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myopic_owl22

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Re: 4U Maths Question Thread
« Reply #2128 on: December 07, 2018, 11:02:22 pm »
0
Hi there!
I was reading through Rui's notes when it's mentioned that the parallelogram method of adding complex number vectors is better than the 'tip to tail' method (which should be "avoided where possible in the MX2 course"). I was wondering what was the reason for this was, given my teacher's strong advocacy for the latter method.
Thanks a bunch!

RuiAce

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Re: 4U Maths Question Thread
« Reply #2129 on: December 07, 2018, 11:53:49 pm »
+4
Hi there!
I was reading through Rui's notes when it's mentioned that the parallelogram method of adding complex number vectors is better than the 'tip to tail' method (which should be "avoided where possible in the MX2 course"). I was wondering what was the reason for this was, given my teacher's strong advocacy for the latter method.
Thanks a bunch!
I'm not a fan of tip-to-tail for half of the complex numbers proofs problems I see. The parallelogram makes it more clear that the main diagonal of the parallelogram is \(z+w\), but then illustrates how the other diagonal represents \(z-w\). It is also the more formal way of doing vector addition.

I've grown to appreciate tip-to-tail for more advanced math problems, because it's just intuitively easier to visualise in comparison (especially when you need more than one vector and thus you risk too many parallelograms). It makes more intuitive sense, so to speak. But the parallelogram has typically been able to give me more useful information in HSC 4U.