-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
complex numbrs (1 Viewer)
- Thread starter sssona09
- Start date
fluffchuck
Active Member
Show that (z^5 - 1)/(z - 1) = z^4 + z^3 + z^2 + z + 1
Start off with your right hand side and consider the sum of a GP.
Also show that (z^5 - 1)/(z - 1) = (z^2 - 2zcos(2pi/5) + 1)(z^2 - 2zcos(4pi/5) + 1)
On the left hand side, we have (z^5 - 1)/(z-1). Try to factorise (z^5 - 1) into the form (z - a)(z - b)(z - c)(z - d)(z - e) where a, b, c, d and e are the zeros of (z^5 - 1) (note: to obtain the zeros of (z^5 - 1), we just set (z^5 - 1) to 0, in other words, z^5 - 1 = 0, which is z^5 = 1, which is root of unity!). From here, you should be able to apply the identity(z-\overline{w}) = z^2 - 2Re(w)z + |w|^2 $$)
, and hence, obtain your right hand side. I hope this helps 
Start off with your right hand side and consider the sum of a GP.
Also show that (z^5 - 1)/(z - 1) = (z^2 - 2zcos(2pi/5) + 1)(z^2 - 2zcos(4pi/5) + 1)
On the left hand side, we have (z^5 - 1)/(z-1). Try to factorise (z^5 - 1) into the form (z - a)(z - b)(z - c)(z - d)(z - e) where a, b, c, d and e are the zeros of (z^5 - 1) (note: to obtain the zeros of (z^5 - 1), we just set (z^5 - 1) to 0, in other words, z^5 - 1 = 0, which is z^5 = 1, which is root of unity!). From here, you should be able to apply the identity
Last edited: