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Re: HSC 2013 4U Marathon $Let $z=(z_1,\ldots,z_n)$ and $w=(w_1,\ldots,w_n)$ be two $n$-tuples of complex numbers. We define:\\ $\langle z,w\rangle:=\sum_{j=1}^n z_j \overline{w_j}.$ \\ I) Prove that $\langle z,w\rangle=\overline{\langle w,z\rangle}$ and $t\langle z,w\rangle=\langle...

I think accuracy is more important than time management at this stage for you. Make sure you understand as much of the material as well as possible, and do questions slowly and carefully. Don't give up on a difficult question easily, bash your head against it for a bit...that is how progress is...

Re: Chester Appreciation Thread!!!!!!!!!!!! Yep, I went there. He and Dr Davyskib were very good if you had an interest in the subject.

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...

Re: HSC 2013 4U Marathon nope.

Re: HSC 2013 4U Marathon Hmmm...I got 2*log(2)-1. Are you sure your answer is correct?

Re: HSC 2013 4U Marathon Well pretty different theme, everything is real here, and the inequality comes directly from squares being positive. But yeah, that is one of the common proofs for the real version of C-S.

Re: HSC 2013 4U Marathon I can't think of how off the top of my head, although there are pretty simple ways to visualise C-S geometrically...in many ways it is more fundamental than the triangle inequality. Was Carrot's question proving C-S for z,w in C^n? Or just in R^n?

Re: HSC 2013 4U Marathon Bingo. Nice result right? I will try to think of harder questions along these lines later.

Re: HSC 2013 4U Marathon At Sy's request a hint: Suppose to the contrary that a root alpha exists in this sector. Evaluating the polynomial P at this alpha we get a sum of (n+1) complex numbers which is supposedly equal to zero. Think about this sum geometrically and a problem should become...

Re: HSC 2013 4U Marathon Yep, implied by the use of inequality signs.

Re: HSC 2013 4U Marathon Another question on polys/complex numbers: $Let $a_k>0$ for $k=0,1,\ldots,n$.\\ Prove that the polynomial\\ \\ $P(z):=\sum_{k=0}^n a_kz^k$ \\ \\has no complex roots $\alpha$ with $|\arg(\alpha)|\leq\pi/n.$ $

Re: HSC 2013 4U Marathon As no-one is answering spirals question I will give a solution for b). The rest is easy. (z+1)^n+(z-1)^n=0\Rightarrow (z+1)^n=-(z-1)^n\Rightarrow |z+1|^n=|z-1|^n\Rightarrow |z+1|=|z-1|. So any roots of this polynomial must lie on the perpendicular bisector of -1 and...

Re: HSC 2013 4U Marathon It does work for 5. Elaborate on your working if you think otherwise, I don't see how you have disproven anything.

$This is just approximating the function $\sqrt{x}$ in an interval by a chord. In this case you are drawing a chord between the points $(\alpha,\sqrt{\alpha})$ and $(\beta,\sqrt{\beta})$ and looking at the y value corresponding to an arbitrary x value on this chord. This can clearly be...

Re: HSC 2013 4U Marathon $Suppose to the contrary that $1+4\cdot 4^{2k}=5p$ for some positive integer $k > 1$ where $p$ is prime. Then\\ \\ $5p=(1+2^{k+1}+2^{2k+1})(1-2^{k+1}+2^{2k+1})$ by difference of squares. \\ \\But for $k>1$ each of these factors is greater than $5$. As the only way to...

Interesting poll results, I expected greater numbers of atheists and agnostics. I suppose it is around the latter part of high school and early part of uni that many people seriously question their faith.

my days haha, makes me feel old. hard to imagine carslaw with benches that aren't wooden, hard and covered in drawings of penii.

Haha yeah the seating was pretty uncomfortable. Mathematicians always get the worst resources though.

Re: HSC 2013 4U Marathon They are roughly equal in length probably.