Post by: gta007 on November 21, 2010, 08:50:24 pm
I stupidly agreed to it, and now have my head hurting for the last half hour.
The question is:
Find all the solutions of (z^3+1)^3=1, where z is a complex number.
I could only get z=0. However she tells me there is 9 SOLUTIONS!!
I think I'm missing something crucial on Le Moivre's theorem.
So was hoping if someone could shed some light on this. Much appreciated. :)
Post by: gta007 on November 21, 2010, 09:16:26 pm
I've worked my way onto..
z=cis(2*pi*k / 9) , k=0,1,2,3,4,5,6,7,8?
Post by: xZero on November 21, 2010, 09:18:16 pm
so Z^3=1, work out 3 solutions
then z^3+1= the 3 solutions and work out 3 more from there
Post by: melissa. on November 21, 2010, 09:19:29 pm
*tries it again*
EDIT:
it gets kind of messy...
but basically i let a = z^3 +1
therefore a^3 = 1 and solved for a
and then i let z^3 + 1 = those solutions of a
therefore z^3 = a - 1
you might need a calculator to convert them back into polar form though
i'd type out my solutions but i have no idea how to use that latex thing
Post by: gta007 on November 21, 2010, 09:28:30 pm
I'll let my friend know, and report back in if there is any further assistance required.
Thanks again. :D
Post by: melissa. on November 21, 2010, 09:29:37 pm
Post by: gta007 on November 21, 2010, 09:35:37 pm
Once again, thanks for the assistance.
Post by: melissa. on November 21, 2010, 09:39:45 pm
Post by: Sunny10 on November 21, 2010, 09:47:12 pm
ahhk cool. if your friend does manage to find all 9 solutions, would you be able to tell us how she found all 3 solutions to z^3 = 0? i'm missing 2 solutions to that...unless i made a mistake earlier on and shouldn't have even got z^3=0
loving the dork in you mel that needs to know all the answers :)
Post by: TrueTears on November 21, 2010, 09:49:22 pm
Clearly has 9 solutions over
Rest is quite trivial although the solution turns out to be quite ugly and I cbf doing the computation but the question is now basically solved.
Post by: TrueTears on November 21, 2010, 09:50:50 pm
You are not wrong, there are 9 solutions if one counts each root up to its multiplicity (http://en.wikipedia.org/wiki/Multiplicity_%28mathematics%29)ahhk cool. if your friend does manage to find all 9 solutions, would you be able to tell us how she found all 3 solutions to z^3 = 0? i'm missing 2 solutions to that...unless i made a mistake earlier on and shouldn't have even got z^3=0
loving the dork in you mel that needs to know all the answers :)
Post by: melissa. on November 21, 2010, 09:56:33 pm
so if you wanted to find the x-intercepts to y=x^3,
x = 0, 0, 0 ... do we take those as three solutions rather than one? (which explains why i'm missing 2 solutions?)
sunny leave me alone :(
Post by: TrueTears on November 21, 2010, 10:00:07 pm
Quote from the wikipedia page:
Quote
The notion of multiplicity is important to be able to count correctly without specifying exceptions (for example, double roots counted twice). Hence the expression, "counted with (sometimes implicit) multiplicity".
The reason we include each root's multiplicity when we use the Fundamental Theorem of Algebra is simply for simplicity's sake. It is like defining 1 to not be a prime number.
Post by: melissa. on November 21, 2010, 10:04:38 pm
Post by: TrueTears on November 21, 2010, 10:08:42 pm
Bahahaha, I saw AzureBlue's karma first and I was like O: until I saw this reply lulzahh, thanks for that. i'd karma you for being "crazy in the maths brain" (sunny's words), but i haven't quite reached the 50 posts milestone yet.No worries, I just did so for you :)
Post by: Sunny10 on November 26, 2010, 05:36:18 pm
ahh, thanks for that. i'd karma you for being "crazy in the maths brain" (sunny's words), but i haven't quite reached the 50 posts milestone yet.
thanks melissa, now he knows we were talking about him, albeit in a very platonic manner :)
Post by: melissa. on November 26, 2010, 09:32:08 pm
Post by: Sunny10 on November 26, 2010, 11:23:28 pm