4U Maths Question Thread

[RuiAce]

Thanks a lot for your help!!! I get it now. Also stuck with part (c) of another question. For these types of questions, I know how to solve the LHS but for the RHS I'm not sure. I tried changing the roots from the equation in part (b) into mod-arg form first and then dividing by z^3 but it didn't work.

Dividing by \(z^3\) before subbing in \(z = \cos\theta+i\sin\theta\) seems to be the required approach here.
\begin{align*}
\frac{RHS}{z^3} &= \frac{(z^2+1)(z^2-\sqrt3 z+1)(z^2+\sqrt3 z + 1)}{z^3}\\
&= \left(\frac{z^2+1}{z} \right) \left(\frac{z^2-\sqrt3z+1}{z}\right)\left(\frac{z^2+\sqrt3z+1}{z}\right)\\
&= \left( z +\frac1z\right) \left(z + \frac1z - \sqrt3\right) \left( z + \frac1z + \sqrt3\right).
\end{align*}
When \(z = \cos\theta+i\sin\theta\), it can be shown that \(z + \frac{1}{z} = 2\cos \theta\). Therefore
\[ \frac{RHS}{z^3} = (2\cos\theta) \left(2\cos\theta - \sqrt3\right) \left(2\cos\theta + \sqrt3\right).  \]
You should be able to finish things from here, noting that \(\sqrt3 = 2\cos\frac\pi6\).

[ursa]

Hello! First time I'm posting on an ATAR Notes forum, so hope I'm doing this right   
In class, we've just been finishing looking at operations on the complex plane. Long story short, I was having difficulty with a question, and the worked solutions provided by the textbook didn't seem to sufficiently answer the question... I spent the whole lesson working with my teacher and the other kid in my ext. 2 class, but we didn't really come up with anything satisfactory.
The question is:
 Prove that the vectors representing the complex numbers u, v and (u - iv)/(1-i) form a right-angled triangle.
Would really appreciate seeing someone's method of solving it! Sorry if my formating is confusing, I couldn't figure out how to attach an image.......
Hope that makes sense, thank you! 

[fun_jirachi]

Welcome to the forums!

In future, you can use an image hosting service like imgur to upload your picture, then copy a link through to the forum to post up a picture. You've definitely got the post right, don't worry about that

I think you seem to have some ideas, so instead of the answer, I'm going to drop a few hints:

Geometric approach
- What does the vector \(\frac{u-iv}{1-i}\) actually denote? Can you express it in terms of some other more familiar/simpler vectors?
- Construct a parallelogram
- Consider the properties of a parallelogram and a kite
- Draw a diagram of all of the above!

Algebraic approach
- Consider the arguments of the vectors \(u-\frac{u-iv}{1-i}\) and \(v-\frac{u-iv}{1-i}\). What do the values of the arguments tell you?

If there's anything else, feel free to ask in this thread 

[vernburn]

Hey there!

Consider what happens if you take \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5)\), where \(\omega = \cos \frac{2\pi}{9} +i\sin \frac{2\pi}{9}\).

We have that \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = (2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9})\).

We also have that by expanding the brackets and simplifying using part (b), \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = -1\).

Hence, we have that \((2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1 \implies \cos \frac{2\pi}{9} \cos \frac{4\pi}{9} \cos \frac{\pi}{9} = 1\).

What's evident in this question is that you occasionally need to look to manipulate the question into a nicer form - here, the solution only really became obvious for me after manipulating the result to \((2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1\). From this sort of position, it's a lot simpler and much less time consuming to explore possible solutions, especially given a number of results from previous parts of the question. Try this line of thinking with similar questions!

Hope this helps

Alternatively, we know that:
\(\begin{aligned}1+z +z ^2+...+z^8 &= \prod_{k=1}^4\left(z- cis\frac{2k\pi}{9}\right)\left(z-cis\left(-\frac{2k\pi}{9}\right)\right)\\
&=\prod_{k=1}^4\left(z^2-2\cos\frac{2k\pi}{9}z+1\right)\end{aligned}\)

letting \(z=i\):
\(\begin{aligned}\prod_{k=0}^4\left(-2i\cos\frac{2k\pi}{9}\right)&=1\\
16\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\cos\frac{2\pi}{3}\cos\frac{8\pi}{9}&=1\\
-\frac{1}{2}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\cos\frac{8\pi}{9}&=\frac{1}{16}\\
-\cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}&=-\frac{1}{8}\\
\implies \cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}&=\frac{1}{8}\end{aligned}\)

[mrsc]

Hi, just wondering what steps you would take to solve these types of questions. I know the graphical method but not sure about the algebraical. Thanks

[Paradoxica]

We can safely assume z²≠1, as the problem itself is undefined if that were true.

Arguments disregard the modulus of the number, so consider the equivalent number upscaled by the positive number The argument of this being equal to π/4 is the same as saying the real and imaginary parts are equal and positive. Hence:This is the cartesian equation of a circle of radius √2 and centre (0,-1).

Now we have to go back and restrict the circle for positive real and imaginary parts.

From the imaginary part -2y, we must have y<0.

Now for the real part, x²+y²>1, we begin by parametrising the circle as This is equivalent to Looking at the parametrisation of we see that this is exactly the same as saying y<0.

Hence, the locus is the arc of the circle with negative imaginary part.

[Husky]

Hey I was wondering if I could get some help with this question from the NESA HSC sample paper for Ext 2. It's question 14 of the paper but It feels harder to understand than the final question, I just can't wrap my head around the strange intervals and use of the pigeonhole principle, even after looking at the solutions.

[fun_jirachi]

Hey there

Imagine you have a number line from 0 to 1 (including 0, not including 1 - the question is kinda dodgy here, since adding one doesn't modify \(\{\alpha\}\)) divvied up into N intervals, each of length 1/N (draw it out!). Counting the numbers they give you, there are N+1 objects and you've drawn up N intervals. You can assign them one by one and there will be an interval that has two numbers in it - since we know that the interval length is 1/N, we can deduce that the difference is less than the length of the interval. Note also that if we put a number at each endpoint of every interval, this would still be the case (as 1, or N/N is excluded - this is why this was key in the first place!).

If you're struggling to visualise it still, I recommend a few things:
- draw it out
- try simpler PHP questions then apply the above to that, see if you can replicate the same ideas here
- come back to it later
- don't worry about it too much, it's only going to stress you out this close to the exam

[Husky]

Hey there

Imagine you have a number line from 0 to 1 (including 0, not including 1 - the question is kinda dodgy here, since adding one doesn't modify \({\alpha}\)) divvied up into N intervals, each of length 1/N (draw it out!). Counting the numbers they give you, there are N+1 objects and you've drawn up N intervals. You can assign them one by one and there will be an interval that has two numbers in it - since we know that the interval length is 1/N, we can deduce that the difference is less than the length of the interval. Note also that if we put a number at each endpoint of every interval, this would still be the case (as 1, or N/N is excluded - this is why this was key in the first place!).

If you're struggling to visualise it still, I recommend a few things:
- draw it out
- try simpler PHP questions then apply the above to that, see if you can replicate the same ideas here
- come back to it later
- don't worry about it too much, it's only going to stress you out this close to the exam

Ah okay that makes much more sense than the NESA solutions which are too brief IMO. Would you mind also explaining the third part of the question?

[fun_jirachi]

Note that as we can represent \(\alpha\) as \(k + \{\alpha\}\). Hence, for some integer n, we can express \(n\alpha\) as \(k + \{n\alpha\}\).
Now, we choose integers a and b in the range [0, N] such that \(\{a\alpha\}\) and \(\{b\alpha\}\) differ by less than 1/N ie. \(0 < |\{a\alpha\} - \{b\alpha\}| < \frac{1}{N}\).

From the above, we can express \(a\alpha\) and \(b\alpha\) as \(k + \{a\alpha\}\) and \(l + \{b\alpha\}\) respectively, where k and l are also integers. We can rearrange to have that \(\{a\alpha\} = a\alpha - k\) and \(\{b\alpha\} = b\alpha - l\), and combining the two, we have that \(\{a\alpha\} - \{b\alpha\} = a\alpha - b\alpha - k + l = (a-b)\alpha - (k - l)\) - but we've noted that a, b, l, and k are integers, and hence subtraction, addition and multiplication using these numbers will definitely result in other integers. Just for the sake of the question, we can let q = a-b and p = k-l, and replace this in the equation I gave in the first paragraph, which matches up with the result given in the question.

When doing questions like this, a few things to note:
- The value of the question in question; it's one mark, so you've definitely already done the hard work beforehand. Look to deduce things from previous parts instead of diving in headfirst. Head first, you will lose as opposed to head first can't lose
- I've reiterated this many times as my teacher reiterated to me; look at what you're working towards. It always helps to have a result in front of you, and it's similar to seeing an answer that's too big, too small, got the wrong sign or the wrong units. Keep yourself on the right track
- Don't stress too much, these are good questions to leave and come back to, especially if you can't see them right away (especially if as you say you find these harder to understand than the questions further back)

Hope this helps

[ti_tirrr]

Hi,

I needed some advice on what to do for my subjects in Year 12. Currently, I have started year 12 term 1 at my school and my subjects are as follows:

4U English
3U Math
Chemistry
Business Studies

I've been content with these subjects and am doing relatively well in them at school. We're starting week 4 at school and I've been thinking; is it too late to consider studying math ext 2? I love doing math but I don't know if I have the right to make that decision yet as Year 12 has already started and so has term classes for math ext 2. My friends in 4U math have already had 2 classes so far. Could anyone give me some advice on what to do about this?

[RuiAce]

Hi,

I needed some advice on what to do for my subjects in Year 12. Currently, I have started year 12 term 1 at my school and my subjects are as follows:

4U English
3U Math
Chemistry
Business Studies

I've been content with these subjects and am doing relatively well in them at school. We're starting week 4 at school and I've been thinking; is it too late to consider studying math ext 2? I love doing math but I don't know if I have the right to make that decision yet as Year 12 has already started and so has term classes for math ext 2. My friends in 4U math have already had 2 classes so far. Could anyone give me some advice on what to do about this?

I don't think it's too late at all. A more important question would be if your teachers would let you. If they deny you entry, that's a barrier out of your control.

On the other hand, catching up is entirely in your control. You might have to deal with a busy 1-3 weeks with the catch-up, but remember that's 1-3 weeks out of around 50 weeks before your HSC exams start. (And remember, you can always drop later if it turned out to be a mistake.)