Search results

just after i finish reading the whole thing i noticed there is a rotated version... why is there a sigma in yours? there is no sum? it's just comparing the term independent of x. on the LHS the term independent of x is just (p+q)C(q)

yeah that's why i mentioned arg(z1) + arg(z2) < pi as the condition before, i noticed that's the only way it worked with z2 in the second quadrant

i did it with z2 in second quadrant and it was still fairly straightforward. if you don't state-rank 4U i will question this world.

i did it with a drawing so not 100% positive. i think it only works if arg(z1)+arg(z2) < pi edit: nope there are more cases than that

oh woops didn't read your example right! uh well i proved it in the exam but i guess there must be an exception. in your example they still lie on the same line that passes through the origin but on opposite sides so the args differ by \pi

also 12b says z1z2 =/= -1 so it is true

did you time yourself?

q^{2}x^{3}-2pqx^{2}+p^{2}x+q or something vaguely similar to that not sure about the signs. i actually didn't get to finish it in the exam :(

also the other thing i didn't understand is why 14.b.i & ii were worth 2 marks each for like 30 seconds of work

i couldn't do ii i understand that is has something to do with |P(i)| but i don't really understand it

oh ffs i tried to do that in the exam but couldn't find a reason why CB/AD = CG/AF man this could have pulled me into the potential high 60s

also 14ci is still wrong T_T should be nx^(n+1) edit: fixed now :)

didn't have time to even reach it :(

you can't just make the assumption that CB/AD = CG/AF can you?

this was by far my favourite question in any exam i have ever sat

question said \frac{(n-1)x^{n} - nx^{n-1} +1}{(x-1)^{2}} had my stumped for a little while

10/10 would volumes again

did not get that far but reading it it looks alright!

even if i got 100 in the externals i would be unlikely to get over 97/100

rereading the multiple choice it amazes me how in exams i can't seem to tell the difference between False and True.