- 28 Jan '04 10:49...prelim time!

i had my AH maths exam yesterday-it was good. a challenge. didn't get 2 questions, and i knew both of them too! one was the wording that got me...[start question]find the 4th roots of unity[end question]. the answer? (0=theta btw) cos0+isin0, cos(0+2pi)+isin(0+2pi), cos(0+4pi)+isin(0+4pi), cos(0+6pi)+isin(0+6pi). de moivres theorem. (well, i think that's the question anyway-but that's definatly the answer...and the other was i had to find the solutions for something along the lines of z^4+5(z^3)+12(z^2)+4z+52. they were all imaginary numbers, and we were already given one of them (2i, i think) so we knew 2 of them (2i, -21) and so we had to find the others. tisn't too hard

but anyway-a kinda clever little question i liked...o.k.-so it's the only other one i remebered, but it's a nice question...(the E is sigma-means to add up everything from k=1 to n...i couldn't find a sigma sign-sorry...)

n

E (4k+1)

k=1

write in terms of n. - 28 Jan '04 16:57the fourth roots of unity are: 1, -1,
*i*, and -*i*, where*i*= sqrt(-1).

the sum can be solved this way:

for k=1, sum = 5;

for k=2, sum = 14 (5+9);

for k=3, sum = 27 (14+13);

for k=4, sum = 44 (27+17).

now use finite differences:

first diff = 9, 13, 17; second diff = 4, 4.

so we have a quadratic equation.

solve it this way:

sum = a*n^2 + b*n + c.

for n=1: a + b + c = 5. (1)

for n=2: 4a + 2b + c = 14. (2)

for n=3: 9a + 3b + c = 27. (3)

subtracting (2) from (3) gives 5a + b = 13. (4)

subtracting (1) from (2) gives 3a + b = 9. (5)

subtracting (5) from (4) gives 2a = 4, so a=2 and b = 3.

substituting shows c to be 0.

the equation is thus: sum((k=1 to n)(4*k+1)) = 2*n^2 + 3*n.

the other equation is too complex for me right now. - 06 Feb '04 18:02 / 1 edit

Factor out the two roots you know, then get the other two by looking at what's left (using the quadratic formula if you can't spot the answer).*Originally posted by genius***...and the other was i had to find the solutions for something along the lines of z^4+5(z^3)+12(z^2)+4z+52. they were all imaginary numbers, and we were already given one of them (2i, i think) so we knew 2 of them (2i,***-2i*) and so we had to find the others. tisn't too hard - 11 Feb '04 22:07

yeah-we went over it in class. tis kinda crap-i scored 65/90. i only just scrpaed an A!and i'd only get 3 questions wrong cause i didn't know how to do them (the two above, and another was the differential of the inverst tanx=1/(1+x). or something like that. it confused me. i'm gonna go over it soon...)-the other 15 to 20 marks were just stupid mistakes. i missread 3 questions, and other silly shizzle like differentialting sin (something), you get (cos (something)).(something)' - but i forgot to change the sin to cos and wrote (sin(something)).(something)' - and in the inverse tan question, i dropped about 5 extra marks cause i simplified the next 3 questions...*rolls eyes* stupid mistakes suck. i*Originally posted by Acolyte***Factor out the two roots you know, then get the other two by looking at what's left (using the quadratic formula if you can't spot the answer).****shall**have got rid of them for my finals! - 19 Feb '04 20:05

the mean score gets you a pass, but the pass line moves so that the SQA can make themselves look better. With last years higher English exam, the pass mark was 35%, and only a third passed the exam. I got a B! this year, we’re doing advanced highers. Kinda harder, and the pass marks are lower. Last years prelim for higher, the A was 80%. (I got 98% ). this year, it was 60%...*Originally posted by richjohnson***That is not uncommon when the instructor sets a difficult exam. It's known as "the curve", reffering to the bell curve of grade distribution. The raw scores are normalized so that the average socres receive C's, the best scores receive A's, etc.** - 22 Feb '04 07:55 / 2 edits

substituting 2*Originally posted by genius*

[b]...the other was i had to find the solutions for something along the lines of z^4+5(z^3)+12(z^2)+4z+52. they were all imaginary numbers, and we were already given one of them (2*i*, i think) so we knew 2 of them (2*i*, -2*i*) and so we had to find the others. tisn't too hard*i*and -2*i*in the equation given does*not*produce a solution as you end up with 20 +/- 32*i*.

i would change the formula to: z^4 + 5*z^3 + 12*z^2 + 20*z + 32 = 0, for which z = 2*i*and z = -2*i*will work.

to get the other two answers from that, divide by z^2 + 4, giving z^2 + 5z + 8, which is easily solved (but not neat).

since i don't know what the true original formula was, i can't say anything else about it.

do you realize how long it's been since i did such an operation???!