HSC 2013 MX2 Marathon (archive) (2 Viewers)

Sy123 · Nov 19, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
$let the points A,B,C and D represent the complex numbers $z_1, z_2, z_3, z_4\\$Since the points are concyclic$, \angle BAD+ \angle ADC=\pi$ (opposite angles in cyclic quadrilaterals are supplementary)$\\\therefore \arg\left ( \frac{z_4-z_1}{z_2-z_1} \right )+\arg\left ( \frac{z_2-z_3}{z_4-z_3} \right )=\pi\\\arg\left ( \frac{(z_4-z_1)(z_2-z_3)}{(z_2-z_1)(z_4-z_3)} \right )=\pi\\ \frac{(z_4-z_1)(z_2-z_3)}{(z_2-z_1)(z_4-z_3)} \right )=-k, (k> 0)\\\therefore$ it is a negative real number$

Nice work.

=========

$x^2-px+q=0 \ \ \ x \in \mathbb{C}$

$$Roots of the quadratic equation are: $ \alpha, \beta$

$S_n=\alpha^n+\beta^n$

$$Prove $ S_{n+2}-pS_{n+1}+qS_n=0$

twinklegal19 · Nov 19, 2012

Re: HSC 2013 4U Marathon

$$Describe the locus of$\\$a) Re$(z^2)+$Im$(z^2)=0\\$b) Arg$(z-1)=\frac{\pi}{4}\\$c) Im$(z)\leq \frac{1}{2}\left | z \right |\\$d) Im\left ( z+\frac{1}{z} \right )=0$

Sy123 · Nov 20, 2012

Re: HSC 2013 4U Marathon

bobmcbob365 said:
a) Find, in modulus-argument form, the roots of the equation z^(2n+1) =1.
b) Hence factorise z^2n + z^(2n-1) + ... + z^2 + z +1 into quadratic factors with real coefficients.
c) Deduce that 2^n * sin(pi/(2n+1)) * sin(2pi/(2n+1)) * sin(3pi/(2n+1)) ... sin(n*pi/(2n+1)) = sqrt(2n+1)

$a) \ \ \ z_0=1cis0, \ \ z_2=1cis \frac{2\pi}{2n+1}, \ \ z_3=1cis\frac{4\pi}{2n+1}..., z_{2n}=1cis\frac{2n\pi}{2n+1}$

$b) \ \ 1+z+z^2+...+z^ {2n}=\frac{z^{2n+1}-1}{z-1}=\frac{(z-1)(z^2-2\cos 2\pi/(2n+1)+1)(z^2-2z\cos \4\pi/(2n+1)+1)...}{(z-1)} \\ \\ = (z^2-2\cos 2\pi/(2n+1)+1(z^2-2\cos 4\pi/(2n+1)+1)...(z^2-2\cos 2\pi n/(2n+1)+1)$

$$Note the identity $ 2-2\cos \theta= 4 \sin^2 \theta/2$

$c) \ \ \ $In the above expression from b, sub in z=1 $ 1+1+1...+1=4^n \sin^2 \pi / (2n+1) \sin^2 2\pi/(2n+1)... \sin^2 n\pi/(2n+1)$

$\sqrt{2n+1}=2^n \sin^2 \pi/(2n+1) \sin^2 2\pi/(2n+1)... \sin^2 n\pi/(2n+1)$

$\square$

HeroicPandas · Nov 20, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Nice work.

=========

$x^2-px+q=0 \ \ \ x \in \mathbb{C}$

$$Roots of the quadratic equation are: $ \alpha, \beta$

$S_n=\alpha^n+\beta^n$

$$Prove $ S_{n+2}-pS_{n+1}+qS_n=0$

let P(x) = x^2-px +q =0

if a and b are roots, therefore P(a) = P(b)=0

P(a) = a^2 -pa+q=0, times by a^n
P(b) = b^2 -pb +q=0, times by b^n

therefore:
a^(n+2) -pa^(n+1) +qa^n =0
b^(n+2) -pb^(n+1) +qb^n =0

solving simultaneously by ADDITION:
a^(n+2) +b^(n+2) -p(a^(n+1) +b^(n+1)) +q(a^n +b^n) =0

S_n =a^n +b^n
S_n+1 =a^(n+1) +b^(n+1)
S_n+2 =a^(n+2) +b^(n+2)

therefore S_n+2 -pS_n+1 + qS_n =0

ALTERNATIVELY,

since a and b are roots of P(x) = 0, then P(a)=P(b)=0

P(a) = a^2 -pa +q =0............[1]
P(b) = b^2 -pb +q =0.............[2]

S_n+2 -pS_n+1 + qS_n =0
a^(n+2) +b^(n+2) -p[a^(n+1) +b^(n+1)] +q(a^n +b^n) =0

a^(n+2) -pa^(n+1) +qa^n +b^(n+2) -pb^(n+1) +qb^n =0

a^n[a^2 -pa +q] + b^n[b^2 -pb +q] =0

by applying [1] and [2]
0+0=0
LHS=RHS therefore, S_n+2 -pS_n+1 + qS_n =0

Sy123 · Nov 20, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
let P(x) = x^2-px +q =0

if a and b are roots, therefore P(a) = P(b)=0

P(a) = a^2 -pa+q=0, times by a^n
P(b) = b^2 -pb +q=0, times by b^n

therefore:
a^(n+2) -pa^(n+1) +qa^n =0
b^(n+2) -pb^(n+1) +qb^n =0

solving simultaneously by ADDITION:
a^(n+2) +b^(n+2) -p(a^(n+1) +b^(n+1)) +q(a^n +b^n) =0

S_n =a^n +b^n
S_n+1 =a^(n+1) +b^(n+1)
S_n+2 =a^(n+2) +b^(n+2)

therefore S_n+2 -pS_n+1 + qS_n =0

ALTERNATIVELY,

since a and b are roots of P(x) = 0, then P(a)=P(b)=0

P(a) = a^2 -pa +q =0............[1]
P(b) = b^2 -pb +q =0.............[2]

S_n+2 -pS_n+1 + qS_n =0
a^(n+2) +b^(n+2) -p[a^(n+1) +b^(n+1)] +q(a^n +b^n) =0

a^(n+2) -pa^(n+1) +qa^n +b^(n+2) -pb^(n+1) +qb^n =0

a^n[a^2 -pa +q] + b^n[b^2 -pb +q] =0

by applying [1] and [2]
0+0=0
LHS=RHS therefore, S_n+2 -pS_n+1 + qS_n =0

Nice work

============

$$Prove that if $ z_1, z_2 $ are complex numbers. Then$ \\ \\ |z_1-z_2|^2+|z_1+z_2|^2=2(|z_1|^2+|z_2|^2)$

twinklegal19 · Nov 20, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Nice work

============

$$Prove that if $ z_1, z_2 $ are complex numbers. Then$ \\ \\ |z_1-z_2|^2+|z_1+z_2|^2=2(|z_1|^2+|z_2|^2)$

$LHS$=|z_1-z_2|^2+|z_1+z_2|^2\\=(z_1-z_2)(\overline{z_1-z_2})+(z_1+z_2)(\overline{z_1+z_2})$ using z\overline{z}=|z|^2$\\=(z_1-z_2)(\overline{z_1}-\overline{z_2})+(z_1+z_2)(\overline{z_1}+\overline{z_2})\\=z_1\overline{z_1}-z_1\overline{z_2}-z_2\overline{z_1}+z_2\overline{z_2}+z_1\overline{z_1}+z_1\overline{z_2}+z_2\overline{z_1}+z_2\overline{z_2}\\=2z_1\overline{z_1}+2z_2\overline{z_2}\\=2(|z_1|^2+|z_2|^2)\\=$RHS$

Sy123 · Nov 20, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
$LHS$=|z_1-z_2|^2+|z_1+z_2|^2\\=(z_1-z_2)(\overline{z_1-z_2})+(z_1+z_2)(\overline{z_1+z_2})$ using z\overline{z}=|z|^2$\\=(z_1-z_2)(\overline{z_1}-\overline{z_2})+(z_1+z_2)(\overline{z_1}+\overline{z_2})\\=z_1\overline{z_1}-z_1\overline{z_2}-z_2\overline{z_1}+z_2\overline{z_2}+z_1\overline{z_1}+z_1\overline{z_2}+z_2\overline{z_1}+z_2\overline{z_2}\\=2z_1\overline{z_1}+2z_2\overline{z_2}\\=2(|z_1|^2+|z_2|^2)\\=$RHS$

Wow haha, not the solution I had in mind it can be done in 3 lines with cosine rule of non-right angled triangles. Either way good insight.

(I would love another one of those really hard questions even if I cant solve it I want to have a go)

HeroicPandas · Nov 20, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Nice work

============

$$Prove that if $ z_1, z_2 $ are complex numbers. Then$ \\ \\ |z_1-z_2|^2+|z_1+z_2|^2=2(|z_1|^2+|z_2|^2)$

dam nvm forgot the indentity zzconj =|z|^2........

Sy123 · Nov 20, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
Is there any more information?

Nope, everything is there. (think vector addition/subtraction or in twinklegal's case identities)

twinklegal19 · Nov 20, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Wow haha, not the solution I had in mind it can be done in 3 lines with cosine rule of non-right angled triangles. Either way good insight.

(I would love another one of those really hard questions even if I cant solve it I want to have a go)

haha, it was the first thing that came into my mind when I saw all those moduli

$$Three complex numbers $z_1,z_2$ and $z_3$ are represented by points $p_1,p_2,p_3$ respectively in an Argand Diagram. If $p_{1}p_{2}p_{3}$ is an equilateral triangle, show that $(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0.$

I think a similar question has been asked before, not sure lol

Sy123 · Nov 20, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
haha, it was the first thing that came into my mind when I saw all those moduli

$$Three complex numbers $z_1,z_2$ and $z_3$ are represented by points $p_1,p_2,p_3$ respectively in an Argand Diagram. If $p_{1}p_{2}p_{3}$ is an equilateral triangle, show that $(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0.$

I think a similar question has been asked before, not sure lol

Yeah Realise asked something similar before and I answered it. I will post a question soon if I can make/find a good one.

twinklegal19 · Nov 20, 2012

Re: HSC 2013 4U Marathon

$\\$a) by writing $-i=i^3$ solve the equation $z^3=i(z+1)^3\\$b) Prove that the roots of the equation $z^3=i(z+1)^3$ are collinear.$$

Sy123 · Nov 20, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
$\\$a) by writing $-i=i^3$ solve the equation $z^3=i(z+1)^3\\$b) Prove that the roots of the equation $z^3=i(z+1)^3$ are collinear.$$

$a)z^3-i(z+1)^3=0 \\ z^3+i^3(z+1)^3=0 \\ \\ (z+i(z+1))(z^2-iz(z+1)-(z+1)^2)=0$

$z=-i(z+1) \Rightarrow z=-iz-i \Rightarrow z(1+i)=-i \\ z_1= \frac{-i}{1+i}\Rightarrow \fbox{z_1=-0.5-0.5i}$

$z^2-iz^2-i-z^2-2z-1=0 \\ \\ iz^2+iz+2z+1=0 \\ \\ z= \frac{-(2+i)\pm \sqrt{(2+i)^2)-4i}}{2i}$

$z_{2,3}=\frac{-2-i \pm \sqrt{3}}{2i}=-0.5+ \frac{-2\pm\sqrt{3}}{2i} \\ \\ \fbox{z_2=-0.5+i(\frac{-2+\sqrt{3}}{2})} \ \ \ \ \ \ \fbox{z_3=-0.5+i(\frac{-2-\sqrt{3}}{2})}$

b)

Re(z_1)=Re(z_2)=Re(z_3)=-0.5

Therefore they all lie on the line x=-0.5 (x=Re)
Therefore collinear

Sy123 · Nov 20, 2012

Re: HSC 2013 4U Marathon

Here is an easy (?) result I just proved.

$S(x)=ax^2+bx+c$

$Prove that if $ S(x) $ has two integer roots. Then b and c is divisible by a$

twinklegal19 · Nov 20, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Here is an easy (?) result I just proved.

$S(x)=ax^2+bx+c$

$Prove that if $ S(x) $ has two integer roots. Then b and c is divisible by a$

not sure if this is right lol. a, b and c are meant to be integers as well right?

$\\S(x)=ax^2+bx+c\\$let the roots be $\alpha$ and $\beta\\\alpha +\beta=-\frac{b}{a}\\b=-a(\alpha +\beta)\\\alpha \beta=\frac{c}{a}\\c=a\alpha \beta\\\therefore$ b and c are divisible by a if $\alpha$ and $\beta$ are integers.$

Sy123 · Nov 20, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
not sure if this is exactly right lol. Are a, b and c are meant to be integers as well? Or just rational/real?

$\\S(x)=ax^2+bx+c\\$let the roots be $\alpha$ and $\beta\\\alpha +\beta=-\frac{b}{a}\\b=-a(\alpha +\beta)\\\alpha \beta=\frac{c}{a}\\c=a\alpha \beta\\\therefore$ b and c are divisible by a if $\alpha$ and $\beta$ are integers.$

Yes, well divisibility implies integers. But you got it right anyway.

================

$i) \ \ \ $By letting $ z=\cos \theta + i\sin \theta $ show that $ z^n+\frac{1}{z^n}=2\cos n\theta$

$ii) \ \ \ $Hence express $ \cos^4 \theta $ in terms of $ \cos k \theta \ \ \ \ k \in \mathbb{N}$

When I say cos k theta, I mean express cos^4 theta as cos theta, cos 2theta, cos 100 theta etc

(Still waiting for a good question)

HeroicPandas · Nov 20, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Yes, well divisibility implies integers. But you got it right anyway.

================

$i) \ \ \ $By letting $ z=\cos \theta + i\sin \theta $ show that $ z^n+\frac{1}{z^n}=2\cos n\theta$

$ii) \ \ \ $Hence express $ \cos^4 \theta $ in terms of $ \cos k \theta \ \ \ \ k \in \mathbb{N}$

When I say cos k theta, I mean express cos^4 theta as cos theta, cos 2theta, etc

(Still waiting for a good question)

i) sub z and by DE MOIVRE'S...
LHS =z^n + z^-n

= cosn(theta) + isinn(theta) + cosn(theta) -isinn(theta) = 2cos ntheta as required

ii) (z+1/z)^4 = z^4 +1/z^4 + 4(z^2+1/z^2) +6

noting z^n +1/z^n = 2cosntheta

threfore, 16cos^4theta = 2cos4theta + 8cos2theta +6

cos^4theta = 1/8 cos4theta + 0.5 cos2theta + 3/8

twinklegal19 · Nov 20, 2012

Re: HSC 2013 4U Marathon

There you go Sy123

$$Suppose $k\left | z-z_1 \right |=l\left | z-z_2 \right |,$ where $k\neq l$ and both are positive real numbers.$\\$(a) Show that the locus of $z$ in the Argand Diagram is a circle with centre $\frac{k^2z_1-l^2z_2}{k^2-l^2}$ and radius $\frac{kl|z_2-z_1|}{|k^2-l^2|}$ by (i) letting $z=x+iy$ (ii) geometric means\\(b)What happens in the limit as $k$ approaches $l?$

HeroicPandas · Nov 20, 2012

Re: HSC 2013 4U Marathon

I) show that (a^2-1)(a^4-14a^2+1) =a^6 - 15a^4 + 15a^2 -1

II) by using de moivre's, show that: cot6theta = [(a^2-1)(a^4-14a^2+1)] / [2a (3a^4 -10a^2 +3)]

where a =cot(theta) and 6theta is not a multiple of PI

III)hence show that cot^2(pi/12) +cot^2(5pi/12) =14

IV)deduce that cot(pi/12) +tan (pi/12) =4

Sy123 · Nov 20, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
i) sub z and by DE MOIVRE'S...
LHS =z^n + z^-n

= cosn(theta) + isinn(theta) + cosn(theta) -isinn(theta) = 2cos ntheta as required

ii) (z+1/z)^4 = z^4 +1/z^4 + 6(z^2+1/z^2) +6

noting z^n +1/z^n = 2cosntheta

threfore, 16cos^4theta = 2cos4theta + 8cos2theta +6

cos^4theta = 1/8 cos4theta + 0.5 cos2theta + 3/8

Correct, good job.

HSC 2013 MX2 Marathon (archive) (2 Viewers)

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