HSC 2013 MX2 Marathon (archive) (3 Viewers)

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bleakarcher

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Re: HSC 2013 4U Marathon

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

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Re: HSC 2013 4U Marathon

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



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



Look at the sum of roots:



Write all of these roots in mod-arg form:



Equate real and imaginary parts we arrive at the result:




b)

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



Apply to the result in part i:








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))


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




Equate co-efficient of x



a=10, a=-9
 

jyu

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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

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

Sy123

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Re: HSC 2013 4U Marathon

b) On our sketch, notice that:





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
Such that it equals to

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

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



 

jyu

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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

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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.
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

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Re: HSC 2013 4U Marathon

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

 
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Re: HSC 2013 4U Marathon


which is equal to one since ? 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 and the denominator is therefore they cancel out?

I need to sleep...
 
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Fus Ro Dah

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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.
Normally would have taken the usual approach to observe that , 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.

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

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

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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

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

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Re: HSC 2013 4U Marathon

Normally would have taken the usual approach to observe that , 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

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Re: HSC 2013 4U Marathon

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

HeroicPandas

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Re: HSC 2013 4U Marathon

b) On our sketch, notice that:





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
Such that it equals to

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

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



very nice
 

jyu

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Re: HSC 2013 4U Marathon

what is very nice? the solution or the question?
 
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