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Re: HSC 2013 4U Marathon Ah yes, my factorisation only works if n is odd. My bad.

Re: HSC 2013 4U Marathon For C, we just do (t+1)(t-cis 2\pi/n)...(t-t_n)=0 t_k (k in N) are the roots. For R, we simply add the conjugates and express as quadratic factors. (t+1)(t^2-2t\cos \frac{2\pi}{n}+1)(t^2-2t\cos \frac{4\pi}{n}+1)... For Q, am I right in saying...

Re: HSC 2013 4U Marathon Is the answer: x=8 x=-1 x=\frac{7\pm i \sqrt{71}}{2}

Re: HSC 2013 4U Marathon $Using the identity $ z^n+\frac{1}{z^n}=2 \cos n\theta $ if $ z= \cos \theta + i \sin \theta $Show that $ \int_0^{\frac{\pi}{2}} \cos^{2n} \theta \ d\theta = \frac{\pi (2n)!}{2^{2n}(n!)^2} Who needs integration by...

Re: HSC 2013 4U Marathon That is what I originally thought, however I was not able to do much with the 3 inequalities that I had and no numbers.

Re: HSC 2013 4U Marathon By Triangle Inequality yes?

Re: HSC 2013 4U Marathon Can the sticks cross over each other, or must the sticks make a perfect triangle without any extra bits sticking out? (Also my slowpoke is more classy (and manly)) than yours :s)

Re: HSC 2013 4U Marathon Well for i, I managed to simply brute force my way through it by: z=x+iy, z_1=a+ib, z_2=c+id Then just doing loads of algebra eventually yielding me the answer. However if this is the only way to do this question (by letting z=x+iy), then I will be a bit...

Re: HSC 2013 4U Marathon Correct, good job.

Re: HSC 2013 4U Marathon 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...

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$

Re: HSC 2013 4U Marathon 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...

Re: HSC 2013 4U Marathon Yeah Realise asked something similar before and I answered it. I will post a question soon if I can make/find a good one.

Re: HSC 2013 4U Marathon Nope, everything is there. (think vector addition/subtraction or in twinklegal's case identities)

Re: HSC 2013 4U Marathon 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)

Re: HSC 2013 4U Marathon 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)

Re: HSC 2013 4U Marathon 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...

Re: HSC 2013 4U Marathon 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

I did not want to say it in the poon thread, but good luck for English.

Re: HSC 2013 4U Marathon Basic but because I cant think of any original questions at the moment and to keep the marathon going: $If the complex numbers $ z_1, z_2, z_3, z_4 $ lie on the Argand plane. Show that if these points form a cyclic quadrilateral then, the complex number...