![]() | |
| | #1 (permalink) |
| New Member HSC: 2009 Gender: Female
Join Date: Sep 2008
Posts: 1
Last Activity:
12 Nov 2009, 7:12 AM ![]() | You can hide this advertisement by registering. Having a bit of trouble with some complex numbers! I was able to do the first bit: Find the roots of z^4+1=0 and show them in an Argand diagram and resolve z^4 into real real quadratic factors.If ya dont understand lol i got the question out of 4unit Patel Mathematics Complex Numers Exercise 4J! Q3 |
| |
| | #2 (permalink) |
| New Member HSC: 2009 Gender: Female
Join Date: Nov 2008
Posts: 14
Last Activity:
19 Nov 2009, 4:20 PM ![]() | Re: Help on question! This is how I did it: z^4 + 1 = (z^2 - 2zcos (pi/4) + 1)(z^2 -2zcos (3pi/4) + 1) Divide both sides by z^2 (for RHS, I just divided each bracket by z, which is the equivalent of dividing by z^2): z^2 + 1/z^2 = (z - 2cos (pi/4) + 1/z)(z - 2cos (3pi/4) + 1/z) Now group z + 1/z and change them to a value of cos@ (z + 1/z = 2cos@ and z^2 + 1/z^2 = 2cos2@): Therefore; 2cos2@ = (2cos@ - 2cos (pi/4))(2cos@ - 2cos (3pi/4)) 2cos2@ = 4 (cos@ - cos (pi/4))(cos@ - cos (3pi/4)) Hence, cos2@ = 2 (cos@ - cos (pi/4))(cos@ - cos (3pi/4)) |
| |
| Bookmarks |
| Thread Tools | |
| Rate This Thread | |
| |
Similar Threads | ||||
| Thread | Thread Starter | Forum | Replies | Last Post |
| Question bout Legislative Power - MC question | SimonLee13 | Law and Society | 2 | 18 Nov 2008 4:04 PM |
| Question for the vegeterians: The Hardest Question a Vegetarian is asked | SomeoneCool | Food & Drinks | 8 | 30 May 2008 5:05 PM |
| How to distinguish btw a Perms/Combs question and a Binomial Probability question? | rsingh | Mathematics (Extension 1) | 4 | 12 Jul 2005 2:41 PM |
| Need help with question 8 for 2003 Economics HSC paper (question inside) | chrome | Economics | 2 | 23 Mar 2004 11:28 PM |
| A Printing Error In Question 8!!! Question Wont Count!!! | Sourie | General Mathematics | 25 | 4 Nov 2002 8:34 PM |