I'm so ****ed I missed out on the last question. I was sitting there thinking I've done everything and 2 minutes before the end I flip through the pages again and see there's a bloody question 10. So annoying especially since it was an easy roots one as well.

Reply 11

SmellyButwifpoop

3

Does anyone still remember the vector question since i forgot i want to retry that to check my original solution

Reply 12

gcw4

11

Original post by studyinkmaths

What did everyone get for the value of 'n' in the very last question... and a and b in the z^10 one

I think from memory for n i got -63, and had no idea for the z^10 one

Reply 13

SmellyButwifpoop

3

can u please tell me the vector question?

Reply 14

SmellyButwifpoop

3

tell me what the question was for z^10 and I can try answer it for you. :smile:

Reply 15

Zo3.

5

Original post by SmellyButwifpoop

tell me what the question was for z^10 and I can try answer it for you. :smile:

oh that question was horrible. I'm pretty sure it was asking for the complex number given:
mod (z^10) = 3^10
arg (z^10) = -5pi/3

Reply 16

harvey_specterx

OP

14

Original post by seadrown

it was pretty easy i'd say

Yeah I actually found it pretty well

Reply 17

harvey_specterx

OP

14

Original post by BartBester

I'm so ****ed I missed out on the last question. I was sitting there thinking I've done everything and 2 minutes before the end I flip through the pages again and see there's a bloody question 10. So annoying especially since it was an easy roots one as well.

Aww that's actually so annoying, wb the rest?

Reply 18

SmellyButwifpoop

3

Original post by Zo3.

oh that question was horrible. I'm pretty sure it was asking for the complex number given:
mod (z^10) = 3^10
arg (z^10) = -5pi/3

Is the question asking what is z? if the answer is yes then here is my solution:
write as mod arg form i.e. 3^10(cos(-5pi/3)+isin(-5pi/3)) we can rewrite this as [3(cos(-1pi/6)+isin(-1pi/6))]^10 since its the same and then we can just consider z = 3(cos(-1pi/6)+isin(-1pi/6)) => z=3sqrt(3)/2 -3i/2, therefore a = 3sqrt(3) all over 2 and b = -3/2