HSC 2012 MX2 Marathon (archive) (1 Viewer)

[IamBread]

Re: 2012 HSC MX2 Marathon

What happened to this thread?

Woo new question XD

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Perhaps you can post up a question now.

 

[cutemouse]

Re: 2012 HSC MX2 Marathon

If 0, z_1, z_2 and z_3 lie on a circle then prove that the points 1/z_1, 1/z_2 and 1/z_3 are collinear. (z_1, z_2, z_3 are non zero complex numbers)

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

If 0, z_1, z_2 and z_3 lie on a circle then prove that the points 1/z_1, 1/z_2 and 1/z_3 are collinear. (z_1, z_2, z_3 are non zero complex numbers)

I've seen this question before... I don't quite remember when.

 

[IamBread]

Re: 2012 HSC MX2 Marathon

Perhaps you can post up a question now.

I couldn't think of any to post
But here's one now lol.

 

[math man]

Re: 2012 HSC MX2 Marathon

I've seen this question before... I don't quite remember when.

question is in patel and terry lee

 

[mnmaa]

Re: 2012 HSC MX2 Marathon

Easy questions
(a).If w is a complex root of z^5 -1=0 with the smallest positive argument, show that w^2 ,w^3 and 2^4 are the other complex roots.And hence prove that 1+w^2+w^3+w^4=0

(b).Find the quadratic equation whose roots are a=w+w^4 and B=w^2+w^3

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Easy questions
(a).If w is a complex root of z^5 -1=0 with the smallest positive argument, show that w^2 ,w^3 and 2^4 are the other complex roots.And hence prove that 1+w^2+w^3+w^4=0

(b).Find the quadratic equation whose roots are a=w+w^4 and B=w^2+w^3

Also, no "Hence find the exact value of":

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

Easy questions
(a).If w is a complex root of z^5 -1=0 with the smallest positive argument, show that w^2 ,w^3 and 2^4 are the other complex roots.And hence prove that 1+w^2+w^3+w^4=0

(b).Find the quadratic equation whose roots are a=w+w^4 and B=w^2+w^3

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ z^5=1=cis0=cis2k\pi\\ z=(cis2k\pi)^{\frac{1}{5}}=cis\frac{2}{5}k\pi_{(k=0,1,2,3,4,)}~\textrm{(By De Moivre's theorem)}\\ ~\\ z_1=cis0=1~~~~~~ z_2=cis\frac{2}{5}\pi~~~~~ z_3=cis\frac{4}{5}\pi~~~~~ z_4=cis\frac{6}{5}\pi~~~~~ z_5=cis\frac{8}{5}\pi\\ ~\\ \therefore w=z_2=cis\frac{2}{5}\pi~\textrm{(Since it's the root with the smallest positive argument)}\\ z_3=(z_2)^2,~~~~~ z_4=(z_2)^3,~~~~~ z_5=(z_2)^4~\textrm{(By De Doivre's theorem)}\\ \therefore \textrm{the other complex roots are}~w^2,w^3, w^4 ~\\~\\~\\~\\ \frac{-b}{a}=\alpha@plus;\beta=w@plus;w^2@plus;w^3@plus;w^4=-1~(\textrm{since}~1@plus;w@plus;w^2@plus;w^3@plus;w^4=0)\\ \therefore b=1\\ ~\\ \frac{c}{a}=\alpha \beta=(w@plus;w^4)(w^2@plus;w^3)\\ =w^3@plus;w^4@plus;w^6@plus;w^7=w^3@plus;w^4@plus;w@plus;w^2=-1~(\textrm{since}~w^5=1)\\ \therefore c=-1\\~\\ \therefore \textrm{the required quadratic equation is}~x^2@plus;x-1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ z^5=1=cis0=cis2k\pi\\ z=(cis2k\pi)^{\frac{1}{5}}=cis\frac{2}{5}k\pi_{(k=0,1,2,3,4,)}~\textrm{(By De Moivre's theorem)}\\ ~\\ z_1=cis0=1~~~~~~ z_2=cis\frac{2}{5}\pi~~~~~ z_3=cis\frac{4}{5}\pi~~~~~ z_4=cis\frac{6}{5}\pi~~~~~ z_5=cis\frac{8}{5}\pi\\ ~\\ \therefore w=z_2=cis\frac{2}{5}\pi~\textrm{(Since it's the root with the smallest positive argument)}\\ z_3=(z_2)^2,~~~~~ z_4=(z_2)^3,~~~~~ z_5=(z_2)^4~\textrm{(By De Doivre's theorem)}\\ \therefore \textrm{the other complex roots are}~w^2,w^3, w^4 ~\\~\\~\\~\\ \frac{-b}{a}=\alpha+\beta=w+w^2+w^3+w^4=-1~(\textrm{since}~1+w+w^2+w^3+w^4=0)\\ \therefore b=1\\ ~\\ \frac{c}{a}=\alpha \beta=(w+w^4)(w^2+w^3)\\ =w^3+w^4+w^6+w^7=w^3+w^4+w+w^2=-1~(\textrm{since}~w^5=1)\\ \therefore c=-1\\~\\ \therefore \textrm{the required quadratic equation is}~x^2+x-1" title="\\ z^5=1=cis0=cis2k\pi\\ z=(cis2k\pi)^{\frac{1}{5}}=cis\frac{2}{5}k\pi_{(k=0,1,2,3,4,)}~\textrm{(By De Moivre's theorem)}\\ ~\\ z_1=cis0=1~~~~~~ z_2=cis\frac{2}{5}\pi~~~~~ z_3=cis\frac{4}{5}\pi~~~~~ z_4=cis\frac{6}{5}\pi~~~~~ z_5=cis\frac{8}{5}\pi\\ ~\\ \therefore w=z_2=cis\frac{2}{5}\pi~\textrm{(Since it's the root with the smallest positive argument)}\\ z_3=(z_2)^2,~~~~~ z_4=(z_2)^3,~~~~~ z_5=(z_2)^4~\textrm{(By De Doivre's theorem)}\\ \therefore \textrm{the other complex roots are}~w^2,w^3, w^4 ~\\~\\~\\~\\ \frac{-b}{a}=\alpha+\beta=w+w^2+w^3+w^4=-1~(\textrm{since}~1+w+w^2+w^3+w^4=0)\\ \therefore b=1\\ ~\\ \frac{c}{a}=\alpha \beta=(w+w^4)(w^2+w^3)\\ =w^3+w^4+w^6+w^7=w^3+w^4+w+w^2=-1~(\textrm{since}~w^5=1)\\ \therefore c=-1\\~\\ \therefore \textrm{the required quadratic equation is}~x^2+x-1" /></a>

 

[AAEldar]

Re: 2012 HSC MX2 Marathon



And for something a little harder...



Interested in seeing different ways people solve it.

 

[zeebobDD]

Re: 2012 HSC MX2 Marathon

And for something a little harder...



Interested in seeing different ways people solve it.

that first one from cambridge polynomials?:/

 

[AAEldar]

Re: 2012 HSC MX2 Marathon

that first one from cambridge polynomials?:/

Yes it is hahaha. Got asked to do it last night and thought it was quite nice.

 

[mnmaa]

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ z^5=1=cis0=cis2k\pi\\ z=(cis2k\pi)^{\frac{1}{5}}=cis\frac{2}{5}k\pi_{(k=0,1,2,3,4,)}~\textrm{(By De Moivre's theorem)}\\ ~\\ z_1=cis0=1~~~~~~ z_2=cis\frac{2}{5}\pi~~~~~ z_3=cis\frac{4}{5}\pi~~~~~ z_4=cis\frac{6}{5}\pi~~~~~ z_5=cis\frac{8}{5}\pi\\ ~\\ \therefore w=z_2=cis\frac{2}{5}\pi~\textrm{(Since it's the root with the smallest positive argument)}\\ z_3=(z_2)^2,~~~~~ z_4=(z_2)^3,~~~~~ z_5=(z_2)^4~\textrm{(By De Doivre's theorem)}\\ \therefore \textrm{the other complex roots are}~w^2,w^3, w^4 ~\\~\\~\\~\\ \frac{-b}{a}=\alpha@plus;\beta=w@plus;w^2@plus;w^3@plus;w^4=-1~(\textrm{since}~1@plus;w@plus;w^2@plus;w^3@plus;w^4=0)\\ \therefore b=1\\ ~\\ \frac{c}{a}=\alpha \beta=(w@plus;w^4)(w^2@plus;w^3)\\ =w^3@plus;w^4@plus;w^6@plus;w^7=w^3@plus;w^4@plus;w@plus;w^2=-1~(\textrm{since}~w^5=1)\\ \therefore c=-1\\~\\ \therefore \textrm{the required quadratic equation is}~x^2@plus;x-1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ z^5=1=cis0=cis2k\pi\\ z=(cis2k\pi)^{\frac{1}{5}}=cis\frac{2}{5}k\pi_{(k=0,1,2,3,4,)}~\textrm{(By De Moivre's theorem)}\\ ~\\ z_1=cis0=1~~~~~~ z_2=cis\frac{2}{5}\pi~~~~~ z_3=cis\frac{4}{5}\pi~~~~~ z_4=cis\frac{6}{5}\pi~~~~~ z_5=cis\frac{8}{5}\pi\\ ~\\ \therefore w=z_2=cis\frac{2}{5}\pi~\textrm{(Since it's the root with the smallest positive argument)}\\ z_3=(z_2)^2,~~~~~ z_4=(z_2)^3,~~~~~ z_5=(z_2)^4~\textrm{(By De Doivre's theorem)}\\ \therefore \textrm{the other complex roots are}~w^2,w^3, w^4 ~\\~\\~\\~\\ \frac{-b}{a}=\alpha+\beta=w+w^2+w^3+w^4=-1~(\textrm{since}~1+w+w^2+w^3+w^4=0)\\ \therefore b=1\\ ~\\ \frac{c}{a}=\alpha \beta=(w+w^4)(w^2+w^3)\\ =w^3+w^4+w^6+w^7=w^3+w^4+w+w^2=-1~(\textrm{since}~w^5=1)\\ \therefore c=-1\\~\\ \therefore \textrm{the required quadratic equation is}~x^2+x-1" title="\\ z^5=1=cis0=cis2k\pi\\ z=(cis2k\pi)^{\frac{1}{5}}=cis\frac{2}{5}k\pi_{(k=0,1,2,3,4,)}~\textrm{(By De Moivre's theorem)}\\ ~\\ z_1=cis0=1~~~~~~ z_2=cis\frac{2}{5}\pi~~~~~ z_3=cis\frac{4}{5}\pi~~~~~ z_4=cis\frac{6}{5}\pi~~~~~ z_5=cis\frac{8}{5}\pi\\ ~\\ \therefore w=z_2=cis\frac{2}{5}\pi~\textrm{(Since it's the root with the smallest positive argument)}\\ z_3=(z_2)^2,~~~~~ z_4=(z_2)^3,~~~~~ z_5=(z_2)^4~\textrm{(By De Doivre's theorem)}\\ \therefore \textrm{the other complex roots are}~w^2,w^3, w^4 ~\\~\\~\\~\\ \frac{-b}{a}=\alpha+\beta=w+w^2+w^3+w^4=-1~(\textrm{since}~1+w+w^2+w^3+w^4=0)\\ \therefore b=1\\ ~\\ \frac{c}{a}=\alpha \beta=(w+w^4)(w^2+w^3)\\ =w^3+w^4+w^6+w^7=w^3+w^4+w+w^2=-1~(\textrm{since}~w^5=1)\\ \therefore c=-1\\~\\ \therefore \textrm{the required quadratic equation is}~x^2+x-1" /></a>

well done but you didnt prove that w+w^2+w^3+w^4=-1

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

First one is straight forward from textbook and HSC paper.

Second one. After substitution plus partialness.

^This? I am leaving the forum if I don't get this answer correct. I've been puzzled by this for a while now. I hope I am not over thinking.

Must be an easier way haha AAEldar.

Note: Sir AAEldar gave me this question a while back.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

^This? I am leaving the forum if I don't get this answer correct. I've been puzzled by this for a while now. I hope I am not over thinking.

You forgot the +C.

Now get out of the forums.

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

You forgot the +C.

Now get out of the forums.

asdogfajogjagrddgs

 

[mnmaa]

Re: 2012 HSC MX2 Marathon

how do you guys "write mathematical notation and symbols on here. I'm currently stuck to using z^2

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

LATTTEEEEXXXXXXX

To start of it's best to use an equation editor until you become fluent.


http://www.codecogs.com/latex/eqneditor.php


Simply add [tex.] to the beginning. And [/tex.] to the end. (no dots.)

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

how do you guys "write mathematical notation and symbols on here. I'm currently stuck to using z^2

http://community.boredofstudies.org/showthread.php?t=234259

 

[mnmaa]

Re: 2012 HSC MX2 Marathon

http://community.boredofstudies.org/showthread.php?t=234259

Thanks brah