OCR Pure Maths 6, stuck on Q

This Question is from January 2005 on Pure Maths 6 for OCR

It is Q 6. iii)

Z donates the complex number 1/2(cos (pie/3) + i sin (pie/3))

Sn donates the sum of the series 1 + z + z^2 +...+z^n-1

Show S6n = (1/3)(1-1/2^6n)(3+i 3^1/2)

Q worth 5 marks, any idea? please help

Reply 1

Jonny W

By the GP formula,

S(6n)
= (1 - z^(6n))/(1 - z)
= (1 - (1/2)^(6n))/(1 - z)

1/(1 - z)
= (1 - z*)/(1 - z - z* + |z|^2)
= (1 - z*)/(1 - cos(pi/3) + 1/4)
= (4/3)(1 - z*)
= (4/3)(1 - (1/2)cos(pi/3) + (1/2)i sin (pi/3))
= (4/3)(3/4 + sqrt(3)i/4)
= (1/3)(3 + sqrt(3)i)

S(6n)
= (1 - z^(6n))/(1 - z)
= (1/3)(1 - (1/2)^(6n))(3 + sqrt(3)i)

Reply 2

Mr_Joe_T

Thanks alot!

I sort of thought that way but got muddled with all the working out, thanks alot though :smile:

Reply 3

naystar

Could you scan in and attach or send me somehow that paper? I'm missing only the P6 Jan '05 one, and I really need the practice!

Reply 4

Mr_Joe_T

i am not great with computers and don't have a scanner, my friend however does, i'll see if he can do it, might take couple of days though

Reply 5

Mr_Joe_T

Hi, i only got the 2nd of 2 pages for P6,i can't upload it as its too big, if you want it email me at [email protected] and ask for it

Reply 6

Mr_Joe_T

I have all of the P6 Jan 2005 paper now, but i can't upload it as it is too big, i changed it to a jpeg but its stil about 280 kb, any ideas how i can make it smaller and get it on here?