HSC 2013 MX2 Marathon (archive) (1 Viewer)

SpiralFlex · Nov 18, 2013

Re: HSC 2013 4U Marathon

For the last Q I posted a hint is the triangular inequality.

RealiseNothing · Nov 18, 2013

Re: HSC 2013 4U Marathon

hit patel said:
haha i feel dumb, . But hope so. What topic is it in?

I was joking about the half yearlies comment haha. It's far beyond HSC maths.

hayabusaboston · Nov 18, 2013

Re: HSC 2013 4U Marathon

I recalled from memory seanieg, naturally I was bound to make an error haha.

Ikki · Nov 18, 2013

Re: HSC 2013 4U Marathon

SpiralFlex said:
For the last Q I posted a hint is the triangular inequality.

Thankyou

Ikki said:
I'm not sure where you got this from but it is a very famous problem:

Fermat's Last Theorem

hit patel said:
Whow How do you show that?> Is that harder ext 1?

I dont get it.

Clearly no one pays attension to the new guy LOL

Ikki · Nov 18, 2013

Re: HSC 2013 4U Marathon

SpiralFlex said:
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$\; \frac{|z|-|y|}{|z-y|} \geq \frac{|z|-|x|}{|z-x|} \geq \frac{|y|-|x|}{|y-x|}$

Triangle Inequality eh spiral? Lets play around with the stuff which is given...

$y=tx+(1-t)z \\ \Indent \therefore t= \frac{y-z}{x-z} \\ \Indent $But we have,$ 0<t<1 \\ \Indent $So by subbing in and multiplying by denom, we get... $ 0<(y-z)<(x-z) \\ \Indent \therefore z<y<x \\ \Indent $Let x,y and z be sides of a triangle where through the inequality proven x is the larger side (Hypotenuse).$ \\ \Indent $From the triangle inequality, we can say that: $ \\ \Indent |z|-|y| \geq |z-y| \\ \Indent \therefore \frac{|z|-|y|}{|z-y|} \geq 1 \\ \Indent $ Similarly, $ \frac{|z|-|x|}{|z-x|} \geq 1 $ and $ \frac{|y|-|x|}{|y-x|} \geq 1 \\ \\ \Indent $And Since, $ z<y<x $ , $ \\ \Indent \therefore \frac{|z|-|y|}{|z-y|} \geq \frac{|z|-|x|}{|z-x|} \geq \frac{|y|-|x|}{|y-x|}$

Idk feels shifty... Did I assume anything? (Can u say those things from triangle ineq?)
Am I on the right path?

seanieg89 · Nov 18, 2013

Re: HSC 2013 4U Marathon

Ikki said:
LOL nice one, caught on technicality? But pretty sure it's referring to fermat's last theorem. Where n must also be an integer.

haha obviously a counter-troll.

Ikki · Nov 19, 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) \$

Thanks for this question

Solution:
$Subbing $(x+iy)$ and Expanding LHS and RHS:$
$x^2+y^2+2y-2xi=7+4i$

$Equate the Imaginary First:$
$-2xi=4i$
$\therefore x=-2$

$Now equate Real and Sub in x=-2$
$x^2+y^2+2y=7$
$y^2+2y-3=0$
$(y+3)(y-1)=0$
$\therefore y=1 $ and $ y=-3$

$\therefore z=-2+i $ or $ z=-2-3i$

More Questions please

hit patel · Nov 19, 2013

Re: HSC 2013 4U Marathon

Ikki said:
Thankyou

Clearly no one pays attension to the new guy LOL

You talking about yourself? I had invaded this forum since term 3. But hey nice to meet you. Its all good hahaha

Ikki · Nov 20, 2013

Re: HSC 2013 4U Marathon

hit patel said:
You talking about yourself? I had invaded this forum since term 3. But hey nice to meet you. Its all good hahaha

Good to meet you too lol. All good bro

hit patel · Nov 20, 2013

Re: HSC 2013 4U Marathon

Ikki said:
Good to meet you too lol. All good bro

Hahaha U asked for question, Here i post one:

Use demoivre theorem to find tan7x in terms of tan x. (3)
Find the Value of cot^2 (pi/7) + cot^2 (2pi/7) + cot^2 (3pi/7). (6)

Note: I didnot make this question and therefore yields correct and valid answers.

Ikki · Nov 21, 2013

Re: HSC 2013 4U Marathon

hit patel said:
Hahaha U asked for question, Here i post one:

Use demoivre theorem to find tan7x in terms of tan x. (3)
Find the Value of cot^2 (pi/7) + cot^2 (2pi/7) + cot^2 (3pi/7). (6)

Note: I didnot make this question and therefore yields correct and valid answers.

Alright, lets do this

[1]
Find tan7x in terms of tanx.

$Using Demoivres: $(cosx+isinx)^7=cos7x+isin7x$

$$ Using normal expansion: $(cosx+isinx)^7=cos^7(x)-21 sin^2(x) cos^5(x)+35 sin^4(x) cos^3(x)+i (-sin^7(x)+7 sin(x) cos^6(x)-35 sin^3(x) cos^4(x)+21 sin^5(x) cos^2(x))-7 sin^6(x) cos(x)$

$\therefore cos7x+isin7x=cos^7(x)-21 sin^2(x) cos^5(x)+35 sin^4(x) cos^3(x)+i (-sin^7(x)+7 sin(x) cos^6(x)-35 sin^3(x) cos^4(x)+21 sin^5(x) cos^2(x))-7 sin^6(x) cos(x)$

$$ Group Reals: $cos7x=cos^7(x)-21 sin^2(x) cos^5(x)+35 sin^4(x) cos^3(x)-7 sin^6(x) cos(x)$

$$ Group Imaginary: $sin7x=-sin^7(x)+7 sin(x) cos^6(x)-35 sin^3(x) cos^4(x)+21 sin^5(x) cos^2(x)$

Divide Imaginary by Real:

$tan7x=\frac{-sin^7(x)+7 sin(x) cos^6(x)-35 sin^3(x) cos^4(x)+21 sin^5(x) cos^2(x)}{cos^7(x)-21 sin^2(x) cos^5(x)+35 sin^4(x) cos^3(x)-7 sin^6(x) cos(x)}$

DIVIDE EVERYTHING BY COS^7(X)

$tan7x=\frac{-tan^7x+7tanx-35tan^3x+21tan^5x}{1-21tan^2x+35tan^4x-7tan^6x}$

YAYYYAYAYAYAYYAYAYYAYAY. Screw the next question, I'll look at it later LOL.

Sy123 · Nov 21, 2013

Re: HSC 2013 4U Marathon

$For what$ \ z \ $is the complex function$ \ \ f(z) = \frac{z+i}{z-i} \ \ $purely imaginary?$

$$Determine the function$ \ g(z) \ $which is purely imaginary for$ \ \ |z| =2$

As for an update on compilation process, I'm up to page 63 right now, its a much longer process than I thought (mostly because I need to make sure the questions are solvable)!

obliviousninja · Nov 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$For what$ \ z \ $is the complex function$ \ \ f(z) = \frac{z+i}{z-i} \ \ $purely imaginary?$

$$Determine the function$ \ g(z) \ $which is purely imaginary for$ \ \ |z| =2$

As for an update on compilation process, I'm up to page 63 right now, its a much longer process than I thought (mostly because I need to make sure the questions are solvable)!

gj Sy.

nightweaver066 · Nov 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$For what$ \ z \ $is the complex function$ \ \ f(z) = \frac{z+i}{z-i} \ \ $purely imaginary?$

$$Determine the function$ \ g(z) \ $which is purely imaginary for$ \ \ |z| =2$

As for an update on compilation process, I'm up to page 63 right now, its a much longer process than I thought (mostly because I need to make sure the questions are solvable)!

Are you sorting by topic too?

Fawun · Nov 21, 2013

Re: HSC 2013 4U Marathon

sigh 4u is going to fuck me up

Ikki · Nov 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$For what$ \ z \ $is the complex function$ \ \ f(z) = \frac{z+i}{z-i} \ \ $purely imaginary?$

$$Determine the function$ \ g(z) \ $which is purely imaginary for$ \ \ |z| =2$

As for an update on compilation process, I'm up to page 63 right now, its a much longer process than I thought (mostly because I need to make sure the questions are solvable)!

What exactly are you compiling?

------
If i let z=x+iy and expand...
$\frac{x^2+y^2-1+2xi}{x^2+(y-1)^2}$

Now logically, if it was to be purely imaginary then, the real part would equal to zero right?

$x^2+y^2-1=0$
$x^2+y^2=1$
$|z|=1$
$z=+-1$

I know i screwed up somewhere...

-------
If its purely imaginary for |z|=2, then.....
$$let $ z=rcistheta$
$z=2cistheta$
Now, let real=0
$costheta=0$
$theta=2npi+-pi/2$

I have a feeling i'm thinking in the wrong way for these questions...

Sy123 · Nov 21, 2013

Re: HSC 2013 4U Marathon

nightweaver066 said:
Are you sorting by topic too?

I can try to do that

Sy123 · Nov 21, 2013

Re: HSC 2013 4U Marathon

Ikki said:
What exactly are you compiling?

------
If i let z=x+iy and expand...
$\frac{x^2+y^2-1+2xi}{x^2+(y-1)^2}$

Now logically, if it was to be purely imaginary then, the real part would equal to zero right?

$x^2+y^2-1=0$
$x^2+y^2=1$
$|z|=1$
$z=+-1$

I know i screwed up somewhere...

-------
If its purely imaginary for |z|=2, then.....
$$let $ z=rcistheta$
$z=2cistheta$
Now, let real=0
$costheta=0$
$theta=2npi+-pi/2$

I have a feeling i'm thinking in the wrong way for these questions...

Yep for the first it is |z| = 1, not sure how you got to z=+- 1 though, z can be complex as well (for example, as an excercise try putting z=1/2 + sqrt(3)/2 i in that function, it should output a purely imaginary number)

For the second one, try thinking of a function similar to the first one.

Ikki · Nov 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yep for the first it is |z| = 1, not sure how you got to z=+- 1 though, z can be complex as well (for example, as an excercise try putting z=1/2 + sqrt(3)/2 i in that function, it should output a purely imaginary number)

For the second one, try thinking of a function similar to the first one.

Jazu... Um... i'm going to go out on a limb and say...
$f(x)=\frac{z-\sqrt(2)i}{z+\sqrt(2)i}$

Well... It seems to work using the previous working out...

Sy123 · Nov 22, 2013

Re: HSC 2013 4U Marathon

Ikki said:
Jazu... Um... i'm going to go out on a limb and say...
$f(x)=\frac{z-\sqrt(2)i}{z+\sqrt(2)i}$

Well... It seems to work using the previous working out...

Replace the sqrt(2) with 2 and its correct

The proper way to do it is just consider:

$f\left(\frac{z}{2} \right) = \frac{\frac{z}{2} +i}{\frac{z}{2} - i}$

We know that it is purely imaginary for

$|\frac{z}{2} | = 1$

Therefore it is purely imaginary for $|z| =2$

HSC 2013 MX2 Marathon (archive) (1 Viewer)

Well-Known Member

what is that?It is Cowpea

Well-Known Member

Active Member

Active Member

Well-Known Member

Active Member

New Member

Active Member

New Member

Active Member

This too shall pass

(╯°□°）╯━︵ ┻━┻ - - - -

Well-Known Member

Queen

Active Member

This too shall pass

This too shall pass

Active Member

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 1)