HSC 2013 MX2 Marathon (archive) (2 Viewers)

RealiseNothing · Nov 21, 2012

Re: HSC 2013 4U Marathon

bleakarcher said:
That's t^n-1.

lol my head's so gone.

bleakarcher · Nov 21, 2012

Re: HSC 2013 4U Marathon

Trebla said:
$$Define $\omega_j=\dfrac{2\pi j}{n}$ for some fixed integers $j$ and $n$. Show that\\\\(a) $\displaystyle\sum_{t=1}^n\cos(t \omega_j) = \displaystyle\sum_{t=1}^n\sin(t \omega_j) = 0\\\\\\$(b) Hence show that for $j\neq k\\\\$(i) $\displaystyle\sum_{t=1}^n\cos(t \omega_j)\cos(t \omega_k) = 0\\\\$(ii) $ \displaystyle\sum_{t=1}^n\sin(t\omega_j)\sin(t \omega_k) = 0\\\\$(iii) $\displaystyle\sum_{t=1}^n\sin(t \omega_j)\cos(t \omega_k) = 0$

a) z+z^2+...+z^n=z[z^n-1]/(z-1)
Let z=cis(w_j).
=> cis(w_j)+[cis(w_j)]^2+...+[cis(w_j)]^n=cis(w_j)[[cis(w_j)]^n-1]/(z-1)
i.e. cis(w_j)+cis(2w_j)+...+cis(nw_j)=cis(w_j)[cis(nw_j)-1]/(z-1) by de Moivre's theorem
=cis(w_j)[cis(n[2pi*j/n])-1]/(z-1)=0
=> sum from t=1 to n [cos(tw_j)]=sum from t=1 to n [sin(tw_j)]=0, equating real and imaginary parts.

Sy123 · Nov 21, 2012

Re: HSC 2013 4U Marathon

Trebla said:
$$Define $\omega_j=\dfrac{2\pi j}{n}$ for some fixed integers $j$ and $n$. Show that\\\\(a) $\displaystyle\sum_{t=1}^n\cos(t \omega_j) = \displaystyle\sum_{t=1}^n\sin(t \omega_j) = 0\\\\\\$(b) Hence show that for $j\neq k\\\\$(i) $\displaystyle\sum_{t=1}^n\cos(t \omega_j)\cos(t \omega_k) = 0\\\\$(ii) $ \displaystyle\sum_{t=1}^n\sin(t\omega_j)\sin(t \omega_k) = 0\\\\$(iii) $\displaystyle\sum_{t=1}^n\sin(t \omega_j)\cos(t \omega_k) = 0$

Note the nth roots of unity. Lets call a general kth one of the nth roots of the unity:

$\alpha_k=\cos \frac{2\pi}{n}k + i \sin \frac{2\pi}{n}k$

If we fix j, we can still put it into alpha and it will still be a root since 'jk' will cover all the possible roots. Hence a general term of the nth roots of unity can be

$\alpha_k=\cos \frac{2\pi j}{n}k+i \sin \frac{2\pi j}{n}k$

Look at the sum of roots:

$\sum_{f=0}^n \alpha_f = 0$

Write all of these roots in mod-arg form:

$\sum_{t=1}^{n} \cos(t\frac{2\pi j}{n})+ i\sum_{t=1}^{n} \sin(t\frac{2\pi j}{n})$

Equate real and imaginary parts we arrive at the result:

$\sum_{t=0}^n \sin(tw_j)= \sum_{t=0}^n \cos(tw_j) = 0$

b)

For part i, note the identity that we can derive:

$0.5(\cos(A+B)+\cos(A-B)=\cos A \cos B$

Apply to the result in part i:

$0.5\sum_{t=1}^n \cos (t(w_j+w_k) + \cos (t(w_j-w_k)$

$w_j+w_k=w_m \ \ \ m \in \mathbb{N}$
$w_k-w_k=w_p \ \ \ p \in \mathbb{N}$

$\Rightarrow 0.5 \sum_{t=1}^n \cos(tw_m)+\cos(tw_p) = 0$

Do things very similar for part ii except with

0.5(cos(A-B)-cos(A+B))

And in part iii

0.5(sin(A+B)+sin(A-B))

Carrotsticks said:
$\\ $Find values of $ a $ and $ b $ such that $ P(x) = (ax+b)(x^5+1)- (5x+1) $ is divisible by $ x^2 + 1$

Is it a=3, b=2? (just need to check before I post my answer)

Carrotsticks said:
$\\ $For what integer value(s) of $ a $ is the polynomial $ P(x) = x^{13} + x + 90 $ divisible by $ x^2 - x + a ?$

$x^{13}+x+90=(x^2-x+a)(x^{11}+x^{10}+...+x^2+x+\frac{90}{a})$

Equate co-efficient of x

$\frac{-90}{a}+a=1$

a=10, a=-9

Sy123 · Nov 22, 2012

Re: HSC 2013 4U Marathon

$z \in \mathbb{C} \ \ \ \ \ |z|=1$

$$Prove $ \ \ \ \ $Re$(\left\frac{1-z}{1+z})\right=0$

jyu · Nov 22, 2012

Re: HSC 2013 4U Marathon

(1-z)(1+z) / (1+z)(1+z)
If |z|=1, the numerator is imaginary, the denominator is real
so the real part of the fraction is zero.

HeroicPandas · Nov 22, 2012

Re: HSC 2013 4U Marathon

jyu said:
(1-z)(1+z) / (1+z)(1+z)
If |z|=1, the numerator is imaginary, the denominator is real
so the real part of the fraction is zero.

Were u chasing for zzconj = |z|^2 to make use of |z|=1?

HeroicPandas · Nov 22, 2012

Re: HSC 2013 4U Marathon

$$P, Q represent complex numbers a, b respectively in an Argand diagram, where O is the origin and O, P, Q are not collinear. In \bigtriangleup OPQ,$ $ the median from O to the mid point M of PQ meets the median from Q to the midpoint N of OP in the point R, where R represents the complex number z. \\\text{a) Show this information on a sketch.} \\\text{b) Explain why there are positive real numbers k, L so that kz = 0.5(a + b)}$ $\text { and L(z-b) = 0.5a - b}$

Sy123 · Nov 22, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
$$P, Q represent complex numbers a, b respectively in an Argand diagram, where O is the origin and O, P, Q are not collinear. In \bigtriangleup OPQ,$ $ the median from O to the mid point M of PQ meets the median from Q to the midpoint N of OP in the point R, where R represents the complex number z. \\\text{a) Show this information on a sketch.} \\\text{b) Explain why there are positive real numbers k, L so that kz = 0.5(a + b)}$ $\text { and L(z-b) = 0.5a - b}$

b) On our sketch, notice that:

$\vec{OM}=0.5(a+b)$

$\vec{QN}=0.5a-b$

Now, the point R lies on both vectors. Indeed for OM, it lies on the vector therefore there must exist some real constant k such that we reduce the modulus of $\vec{OM}$
Such that it equals to $\vec{OR}=z$

Similarly, QN is \vec{ON}-b. And R lies on this line, hence there exists some constant L such that we reduce the modulus by such a factor to make L(z-b)=0.5a-b

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

$i) $Given that $ \ \ z=\cos \theta + i\sin \theta \ \ $Show that$ \ \ z^n + z^{-n} \ \ $ is real for positive integers n$$

$ii) $Hence or otherwise solve the equation$ \ \ \ 2z^4+3z^3+5z^2+3z+2=0$

jyu · Nov 22, 2012

Re: HSC 2013 4U Marathon

u, v and w are the roots of (sqrt2)z^3 = 1+i. Show that (1+u)/(1-u), (1+v)/(1-v) and (1+w)/(1-w) are collinear.

Sy123 · Nov 22, 2012

Re: HSC 2013 4U Marathon

jyu said:
u, v and w are the roots of (sqrt2)z^3 = 1+i. Show that (1+u)/(1-u), (1+v)/(1-v) and (1+w)/(1-w) are collinear.

Lets cheat and use the result from the previous question I asked.

Although they are reciprocals of the question I asked, the result still applies.
We can prove that if they have modulus 1 (which u, v, w all do when solving the equation), then they are all purely imaginary, hence they all are collinear.

Arr0gance · Nov 22, 2012

Re: HSC 2013 4U Marathon

Hey guys, a short question. Pretty simple but i really liked it

$$ If $ \alpha , \beta $ are two complex numbers where $ \alpha\neq\beta $ and $ |\alpha|=1, $ show that $ \left |\frac {\alpha\bar{\beta}-1}{\alpha-\beta} \right |=1$

twinklegal19 · Nov 22, 2012

Re: HSC 2013 4U Marathon

$\left | \frac{\alpha \overline{\beta}-1}{\alpha-\beta} \right |\\\\=\frac{|\alpha \overline{\beta}-1|}{|\alpha-\beta|}\\\\=\frac{|\alpha\overline{\beta}-1|}{|1-\frac{\beta}{\alpha}||\alpha|}\\\\$since $|\alpha|=1,\alpha^{-1}=\overline{\alpha}\\\\=\frac{|\alpha\overline{\beta}-1|}{|1-\beta\overline{\alpha}|}\\\\=\frac{|\alpha\overline{\beta}-1|}{|\beta\overline{\alpha}-1|}$
which is equal to one ~~since $|\overline{z}|=|z|$ ? this is probably the wrong reasoning lol, do you have to split it up using the triangular inequality or something?~~

edit: can you say the numerator is $\geq |\alpha\overline{\beta}|-1$ and the denominator is $\geq |\beta\overline{\alpha}|-1$ therefore they cancel out?

I need to sleep...

Fus Ro Dah · Nov 23, 2012

Re: HSC 2013 4U Marathon

jyu said:
u, v and w are the roots of (sqrt2)z^3 = 1+i. Show that (1+u)/(1-u), (1+v)/(1-v) and (1+w)/(1-w) are collinear.

Normally would have taken the usual $z' \mapsto \frac{1+z}{1-z}$$ approach to observe that $z = \frac{z'-1}{z'+1}$ , and then substitute into P(z). We acquire a polynomial that is easy to prove has all complex roots but proving that they are purely imaginary is so long and tedious, may as well just solve P(z) and substitute all the values in. I can't immediately think of a shorter method at the moment.

Arr0gance said:
Hey guys, a short question. Pretty simple but i really liked it

$$ If $ \alpha , \beta $ are two complex numbers where $ \alpha\neq\beta $ and $ |\alpha|=1, $ show that $ \left |\frac {\alpha\bar{\beta}-1}{\alpha-\beta} \right |=1$

Construct vectors A, B, B*, AB* and AB*-1. We immediately observe two congruent triangles, where matching sides include |A-B| and |AB*-1| and hence the result.

seanieg89 · Nov 23, 2012

Re: HSC 2013 4U Marathon

For jyu's question the polynomial equation is equivalent to taking the cube roots of a complex number of unit modulus. These three roots must lie on the unit circle.

But the map $z\mapsto \frac{1+z}{1-z}$

sends the unit circle (excluding 1) to the imaginary axis. To see this, simply multiply the numerator and denominator of this fraction by the conjugate of the denominator and simplify.

This implies the the images of u,v,w under this transformation all lie on the imaginary axis.

Carrotsticks · Nov 23, 2012

Re: HSC 2013 4U Marathon

Fus Ro Dah said:
Normally would have taken the usual $z' \mapsto \frac{1+z}{1-z}$$ approach to observe that $z = \frac{z'-1}{z'+1}$ , and then substitute into P(z). We acquire a polynomial that is easy to prove has all complex roots but proving that they are purely imaginary is so long and tedious, may as well just solve P(z) and substitute all the values in. I can't immediately think of a shorter method at the moment.

Might want to read up on this.

http://en.wikipedia.org/wiki/Möbius_transformation

cutemouse · Nov 23, 2012

Re: HSC 2013 4U Marathon

Another important application of this is the 'bilinear transformation' in digital signal processing.

HeroicPandas · Nov 23, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
b) On our sketch, notice that:

$\vec{OM}=0.5(a+b)$

$\vec{QN}=0.5a-b$

Now, the point R lies on both vectors. Indeed for OM, it lies on the vector therefore there must exist some real constant k such that we reduce the modulus of $\vec{OM}$
Such that it equals to $\vec{OR}=z$

Similarly, QN is \vec{ON}-b. And R lies on this line, hence there exists some constant L such that we reduce the modulus by such a factor to make L(z-b)=0.5a-b

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

$i) $Given that $ \ \ z=\cos \theta + i\sin \theta \ \ $Show that$ \ \ z^n + z^{-n} \ \ $ is real for positive integers n$$

$ii) $Hence or otherwise solve the equation$ \ \ \ 2z^4+3z^3+5z^2+3z+2=0$

very nice

jyu · Nov 24, 2012

Re: HSC 2013 4U Marathon

what is very nice? the solution or the question?

twinklegal19 · Nov 24, 2012

Re: HSC 2013 4U Marathon

probably both

HeroicPandas · Nov 24, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
probably both

Yes

HSC 2013 MX2 Marathon (archive) (2 Viewers)

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