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Just seems a bit silly to assume something like those points being concyclic. From that assumption we can also prove things like: -any two real numbers are equal -triangles have angle sum 1,000,000 -any two functions are the same

Spiral is correct, the angle will be less than 90 because of the fallacious premise I mentioned.

Yeah generally a complete proof will utilise something relatively deep...such as the Jordan curve theorem from topology or Liouville's theorem from complex analysis. This is the first step in several proofs though I believe. PS I do recall seeing a relatively elementary proof of FTA somewhere...

Complex analysis is indeed magical. If any of you are interested in number theory you should check out how complex analysis is used to prove the prime number theorem (eg the second last chapter of Tom Apostol's Introduction to Analytic Number Theory)...it is pretty mindblowing the first time you...

I did not mean any part in particular. I simply meant that the premise of the question is always false. The four points corresponding to 0, z_2, z_3, z_1+z_2 CANNOT ever be concyclic. (Try to find z_1 for which they are...I can provide geometric justification for my claim if you wish). So the...

Merry xmas BOS! Umm, I am reasonably sure the geometric situation in mathmans original question is not actually possible, and hence the identity required to prove should never actually hold...maybe I am mistaken. Here is another question, based on the triangle inequality. It is commonly used...

Seen PWD live ~5 times ;). Yes you are haha.

you have excellent taste in music :).

Haha I did the same thing.

Going back a bit, but 1993 was quite hard. Haven't looked at it recently enough to compare to 2011, but it is probably of similar difficulty, I remember the Coroneos solution to Q8 was badly wrong!

Well it is not entirely clear how Carrotsticks defines the family of lines in his diagram. All that is indicated is that the x-intercepts are equally spaced. If the lines are such that the y-intercepts are also equally spaced, then the curve you wrote down first is correct. If the lines are...

OK, it seems the different answers stem from the fact that you two used a different definition for the family of defining lines to me. I took it to be the set of line segments from the x-axis to the y-axis of length 1. (Think of the side-on view of a ladder sliding down from a wall with its...

$Okay I showed that the curve is an astroid. \\Let $\ell_t$ denote the line segment joining the points $(t,0)$ and $(0,\sqrt{1-t^2}).$ Any point \emph{above} the limiting curve in question must lie above $\ell_t$ for every $t\in (0,1).$ \\Moreover, the limiting curve is the graph of the function...

Got R=1/6 by just taking the limit as n->inf of the area enclosed by the envelope with K=1/n (it is fairly straightforward to calculate). My method should also yield the equation of the curve by using the FTOC (we can calculate partial areas in a similar way and hence obtain a differential...

I for one haven't drawn that thing before :P, how is it defined? Edit: Oh wait, I think I see what you mean. Going to try to figure it out :)

z=2\textrm{cis}\left(\frac{-\pi+4k\pi}{8}\right)\quad 0\leq k <4

multiplying out i'm guessing, its not like the difference in speed is substantial...

1. The condition given implies that the parallelogram spanned by z_1 and z_2 is a rhombus (perpendicular diagonals) with the z_1+z_2 diagonal having length twice that of the diagonal z_1-z_2. (Also make sure the orientation of your rhombus is correct.) c) is simple trig once you understand the...

$Alternatively:$ \\z-w=2(\textrm{cis}(\pi/2)-\textrm{cis}(2\pi/3))\\=2\textrm{cis}(7\pi/12)\cdot(\textrm{cis}(-\pi/12)-\textrm{cis}(\pi/12))=-4i\sin(\pi/12)\textrm{cis}(7\pi/12).\\ \\$Hence: $\arg(z-w)=\arg(-i)+\arg(\textrm{cis}(7\pi/12))=7\pi/12-\pi/2=\pi/12.

$Yet another solution using De Moivre:\\ \\$ \sum_{k=0}^{n-1}\textrm{cis}(k\theta)=\frac{z^n-1}{z-1}=\frac{z^{n/2}(z^{n/2}-z^{-n/2})}{z^{1/2}(z^{1/2}-z^{-1/2})}=z^{\frac{n-1}{2}}\frac{2i\sin(n\theta/2)}{2i\sin(\theta/2)}\\ =\textrm{cis}((n-1)\theta/2)\frac{\sin(n\theta/2)}{\sin(\theta/2)}.\\ \\...