SOLUTION: i) Expand z = {{{(1+ic)^6}}} in powers of c ii) Hence find the 5 real values of c for which z is real.

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: i) Expand z = {{{(1+ic)^6}}} in powers of c ii) Hence find the 5 real values of c for which z is real.      Log On


   

Question 1152858: i) Expand z = %281%2Bic%29%5E6 in powers of c
ii) Hence find the 5 real values of c for which z is real.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


%281%2Bic%29%5E6



1+%2B+6ic+-+15c%5E2+-+20ic%5E3+%2B+15c%5E4+%2B+6ic%5E5+-+c%5E6

%281-15c%5E2%2B15c%5E4%29+%2B+%286c-20c%5E3%2B6c%5E5%29i

We want the imaginary part to be 0:

6c-20c%5E3%2B6c%5E5+=+0
2c%283-10c%5E2%2B3c%5E4%29+=+0
2c%283-c%5E2%29%281-3c%5E2%29+=+0

c=+0 OR c%5E2+=+3 OR c%5E2+=+1%2F3

The 5 values of c for which the expression is real are
0
sqrt%283%29 and -sqrt%283%29
sqrt%281%2F3%29 and -sqrt%281%2F3%29