# Thread: 2012 HSC MX2 Marathon

1. ## Re: 2012 HSC MX2 Marathon

centre: (1,√3)
√3.

$\\\textup{Let }z_{1}=3\left (\cos\left (\frac{\pi}{3} \right )+i\sin\left (\frac{\pi}{3} \right )\right ) \textup{ and } z_{2}=4\left (\cos\left (\frac{\pi}{6} \right )+i\sin\left (\frac{\pi}{6} \right )\right ).\\\\\textup{Determine the smallest integral } n \textup{ for which } \left ( \frac{z_{1}}{z_{2}} \right )^{n} \textup{ is real}.$

2. ## Re: 2012 HSC MX2 Marathon

n=6 (assuming n=0 isn't allowed).

Given that w is the complex root of z^3=1 with smallest positive argument, evaluate:
(1-w)(1-w^2)(1-w^4)(1-w^8)

3. ## Re: 2012 HSC MX2 Marathon

Also smallest positive integral value.

4. ## Re: 2012 HSC MX2 Marathon

Level I

Level II

Level III

TBA tomorrow when I am bothered.

5. ## Re: 2012 HSC MX2 Marathon

Originally Posted by SpiralFlex
Level I

Level II

Level III

TBA tomorrow when I am bothered.
This is a good idea. Having difficulty 'levels' when posting up questions is a good way to cater for different skill levels on the forums.

However, I suggest you co-ordinate the colours with the level of difficulty.

ie: Green = Easy, Blue = Medium, Red = Hard.

Black = Destroy.

6. ## Re: 2012 HSC MX2 Marathon

Originally Posted by Carrotsticks
This is a good idea. Having difficulty 'levels' when posting up questions is a good way to cater for different skill levels on the forums.

However, I suggest you co-ordinate the colours with the level of difficulty.

ie: Green = Easy, Blue = Medium, Red = Hard.

Black = Destroy.
It's always black when it comes to negative. =.=

7. ## Re: 2012 HSC MX2 Marathon

^I know right...why can't it be white?

8. ## Re: 2012 HSC MX2 Marathon

Originally Posted by bleakarcher
^I know right...why can't it be white?

9. ## Re: 2012 HSC MX2 Marathon

Originally Posted by Carrotsticks
This is a bit of a tricky question.

However looking at the quality of the questions and solutions so far, I'm quite convinced that one of ther 2012'ers will get it:

Preuni question?

10. ## Re: 2012 HSC MX2 Marathon

Originally Posted by nightweaver066
Preuni question?
Do you go to Pre Uni?

11. ## Re: 2012 HSC MX2 Marathon

Originally Posted by Carrotsticks
Do you go to Pre Uni?
Nope.

12. ## Re: 2012 HSC MX2 Marathon

Guess I'll contribute a question as well..

$Using ln(z) = ln|z| + iarg(z)$

$Find i^\frac{1}{i}$

13. ## Re: 2012 HSC MX2 Marathon

Originally Posted by Carrotsticks
The 'little circle' is the universal symbol for 'function composition'.
Hahahaha. That made LOL. Owned him :P

14. ## Re: 2012 HSC MX2 Marathon

Guess I'll contribute a question as well..

$Using ln(z) = ln|z| + iarg(z)$

$Find i^\frac{1}{i}$
$i^\frac{1}{i}$

$= e^{\frac{1}{i}ln(i)}$

$Let z = i$

$\therefore ln(i) = ln|i| + iarg(i)$

$= i\frac{\pi}{2}$

$\therefore i^\frac{1}{i} = e^{\frac{1}{i} \times i\frac{\pi}{2}$

$= e^\frac{\pi}{2}$

Question

15. ## Re: 2012 HSC MX2 Marathon

Originally Posted by deswa1
n=6 (assuming n=0 isn't allowed).

Given that w is the complex root of z^3=1 with smallest positive argument, evaluate:
(1-w)(1-w^2)(1-w^4)(1-w^8)
= (1 - w)(1 - w^2)(1 - w)(1 - w^2)

= (1 - 2w + w^2)(1 - 2w^2 + w^4)

= (-3w)(1 + w - 2w^2)

= (-3w)(-3w^2)

= 9w^3

= 9

16. ## Re: 2012 HSC MX2 Marathon

Originally Posted by nightweaver066
$i^\frac{1}{i}$

$= e^{\frac{1}{i}ln(i)}$

$Let z = i$

$\therefore ln(i) = ln|i| + iarg(i)$

$= i\frac{\pi}{2}$

$\therefore i^\frac{1}{i} = e^{\frac{1}{i} \times i\frac{\pi}{2}$

$= e^\frac{\pi}{2}$

Question
Is this a past HSC question? I remember doing this exact question about two days ago. I'll let someone else post a solution though.

With your previous post, the answer is a nice integer. You made a mistake in the second line where you changed (1-w)(1-w)(1-w^2)^2 into (1-w^2)^3

17. ## Re: 2012 HSC MX2 Marathon

yeh its 9.

18. Originally Posted by bleakarcher
yeh its 9.
No, it's 8.
No, it's 9.
Yes, that's right, I was testing you. It's 9.

19. ## Re: 2012 HSC MX2 Marathon

Originally Posted by Riproot
No, it's 8.
No, it's 9.
Yes, that's right, I was testing you. It's 9.
And that's a magic number.....

20. ## Re: 2012 HSC MX2 Marathon

Originally Posted by deswa1
Is this a past HSC question? I remember doing this exact question about two days ago. I'll let someone else post a solution though.

With your previous post, the answer is a nice integer. You made a mistake in the second line where you changed (1-w)(1-w)(1-w^2)^2 into (1-w^2)^3
Yeah it is a past HSC question.

Thanks. Got too excited looking for a difference of two squares.

21. ## Re: 2012 HSC MX2 Marathon

Originally Posted by bleakarcher
yeh its 9.
What's 9?

22. ## Re: 2012 HSC MX2 Marathon

The answer to my question, a few posts back

23. ## Re: 2012 HSC MX2 Marathon

Originally Posted by nightweaver066
Question

5 minutes to work out. 1 hour to type up. 30 minutes to post. I need to get better at latex.

24. ## Re: 2012 HSC MX2 Marathon

Originally Posted by Nooblet94

5 minutes to work out. 1 hour to type up. 30 minutes to post. I need to get better at latex.
1 hour's good by my standards for that length. It took me 25 minutes to type up a solution to the fourth roots of unity (about 4 lines of working). I've quit latex now...

25. ## Re: 2012 HSC MX2 Marathon

Originally Posted by deswa1
1 hour's good by my standards for that length. It took me 25 minutes to type up a solution to the fourth roots of unity (about 4 lines of working). I've quit latex now...
Haha yeah latex takes a bit to get used too...

$If sinx = \frac{e^{ix}-e^{-ix}}{2i} , show sin^{-1}x = -i\ln(ix+\sqrt{1-x^2})$

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