HSC 2012 MX2 Marathon (archive) (3 Viewers)

rolpsy · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

centre: (1,√3)
radius: 2√3.

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

deswa1 · Jan 14, 2012

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)

SpiralFlex · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Also smallest positive integral value.

SpiralFlex · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Level I

Level II

Level III

TBA tomorrow when I am bothered.

Carrotsticks · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
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.

SpiralFlex · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
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. =.=

bleakarcher · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

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

Carrotsticks · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

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

Because nobody can read white.

nightweaver066 · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
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?

Carrotsticks · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
Preuni question?

Do you go to Pre Uni?

nightweaver066 · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Do you go to Pre Uni?

Nope.

IamBread · Jan 15, 2012

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}$

cutemouse · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
The 'little circle' is the universal symbol for 'function composition'.

Hahahaha. That made LOL. Owned him

nightweaver066 · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
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

nightweaver066 · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
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

deswa1 · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
$i^\frac{1}{i}$

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

$$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

bleakarcher · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

yeh its 9.

Riproot · Jan 15, 2012

bleakarcher said:
yeh its 9.

No, it's 8.
No, it's 9.
Yes, that's right, I was testing you. It's 9.

Carrotsticks · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

Riproot said:
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.....

nightweaver066 · Jan 15, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
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.

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