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.
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.
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