HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Oct 25, 2013

Re: HSC 2013 4U Marathon

I can gradually latex it all up in a pdf

SpiralFlex · Oct 25, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
I can gradually latex it all up in a pdf

97 pages, will take longer than anticipated from experience of latexing assignments (assuming you want to align everything correctly). But hey you got 3 month break.

Sy123 · Oct 25, 2013

Re: HSC 2013 4U Marathon

SpiralFlex said:
97 pages, will take longer than anticipated from experience of latexing assignments (assuming you want to align everything correctly). But hey you got 3 month break.

Yea lol, though the latex won't be professional at all so that might make it faster

Edit: Already up to page 10, I'll continue later
Edit2: Nice way to pass the time, up to page 22
Edit3: Page 41
Edit4: Page 56

hit patel · Oct 26, 2013

Re: HSC 2013 4U Marathon

SpiralFlex said:
I think it would be good if someone if bothered can compile all the questions that were posted here in a pdf and make and give it to the 2014ers in a 4U marathon 2014 thread and continue the tradition?

+1 spiral

But It would be an overload task so yeh.

hit patel · Oct 26, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yea lol, though the latex won't be professional at all so that might make it faster

Edit: Already up to page 10, I'll continue later
Edit2: Nice way to pass the time, up to page 22

Oh ok. cool i will do this next year for the 2015'ers.

HAX0R · Nov 7, 2013

Re: HSC 2013 4U Marathon

How's it going?

Sy123 · Nov 7, 2013

Re: HSC 2013 4U Marathon

HAX0R said:
How's it going?

I took a little (a big) break with it, I'll make sure to get to at least page 40 by tonight and continue regularly from there

Fizzy_Cyst · Nov 10, 2013

Re: HSC 2013 4U Marathon

New Question:

$$Find all the complex numbers$ \ z=a+ib, \ $where a and b are real, such that$ \ |z|^2 - 7 = 2i(z+2) \$

braintic · Nov 10, 2013

Re: HSC 2013 4U Marathon

Fizzy_Cyst said:
New Question:

$$Find all the complex numbers$ \ z=a+ib, \ $where a and b are real, such that$ \ |z|^2 - 7 = 2i(z+2) \$

What is the significance of the dollar sign? I've never seen complex numbers applied to financial maths before.

Fizzy_Cyst · Nov 10, 2013

Re: HSC 2013 4U Marathon

Its my noob latex skillllllz

Kurosaki · Nov 10, 2013

Re: HSC 2013 4U Marathon

Fizzy_Cyst said:
New Question:

$$Find all the complex numbers$ \ z=a+ib, \ $where a and b are real, such that$ \ |z|^2 - 7 = 2i(z+2) \$

z=-2-3i and z=-2+i?

Kurosaki · Nov 10, 2013

Re: HSC 2013 4U Marathon

I did it like this- $|z|^2=a^2+b^2$
then LHS= $a^2+b^2-7$ .
RHS= $2i(a+ib+2)=2ai-2b+4i=-2b+i(2a+4)$
equating coefficients, we see that there is no imaginary part on the LHS, since a and b are real, hence a=-2.
We then substitute that into the LHS and RHS, to find that
$b^2-3=-2b$
Solving for b, we get b=1 and b=-3?

Fizzy_Cyst · Nov 10, 2013

Re: HSC 2013 4U Marathon

Good stuff!

$$Find the square roots of$ \ 1 + 2 \sqrt{2}i$

anomalousdecay · Nov 10, 2013

Re: HSC 2013 4U Marathon

Find the cubic roots of:

$w^3 = 1$

in (i) mod-arg form and (ii) the form x+iy, where x and y are real.
I'm not sure if this is too early in the year to post for 2014'ers.

EDIT: By the way, there are two ways to do this question.

Carrotsticks · Nov 10, 2013

Re: HSC 2013 4U Marathon

anomalousdecay said:
Find the roots of the unity of:

$w^3$

I'm not sure if this is too early in the year to post for 2014'ers.

That is not a valid question.

braintic · Nov 10, 2013

Re: HSC 2013 4U Marathon

anomalousdecay said:
Find the roots of the unity of:

$w^3$

I'm not sure if this is too early in the year to post for 2014'ers.

Not only not valid - I can even begin to place an interpretation on it.

anomalousdecay · Nov 10, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
That is not a valid question.

Doesn't "of the unity" mean it equals one though?
And I know one mistake I did was I forgot to put in "cubic"

i.e.

$w^3 =1$

anomalousdecay · Nov 10, 2013

Re: HSC 2013 4U Marathon

braintic said:
Not only not valid - I can even begin to place an interpretation on it.

what do you mean?

SpiralFlex · Nov 10, 2013

Re: HSC 2013 4U Marathon

Put a new 2014 thread? I dislike years getting mixed up =[

$\fbox{1}\; $Solve the equation$\; z^2-8(1-i)z+63-16i=0$

$\fbox{2}\; $Prove the parallelogram law$\; |z_{1}-z_{2}|^2+|z_{1}+z_{2}|^2 = 2(|z_{1}|^2+|z_{2}|^2) \;$for all complex numbers$\;z_{1}, z_{2}. \;$You may want to consider a parallelogram in$\; \mathbb{C} \;$with the vertices$\; \mathbf{0}, z_{1}, z_{2}, z_{1}+z_{2}$

$\fbox{3*}\; $Let$\; x, y, z \;$be distinct complex numbers such that$\; y=tx+(1-t)z \;$where$\; 0<t<1.$

$$Prove that$\; \frac{|z|-|y|}{|z-y|} \geq \frac{|z|-|x|}{|z-x|} \geq \frac{|y|-|x|}{|y-x|}$

braintic · Nov 10, 2013

Re: HSC 2013 4U Marathon

anomalousdecay said:
what do you mean?

You've edited the question since I made my post. The original question made no sense.

As it stands now, I know what you mean, but the wording is confusing. You just need to say 'find the complex cube roots of 1'.

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