Complex number question.

[gta007]

Okay guys and girls, I have been asked by a friend in NZ to help her with a question.
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. 

[gta007]

Hmmmm...
I've worked my way onto..
z=cis(2*pi*k / 9) , k=0,1,2,3,4,5,6,7,8?

[xZero]

i would treat z^3+1 as Z, work out 3 solutions then sub z^3+1 back in and get 3 more solutions in each solution so u end up with 3x3=9 solutions

so Z^3=1, work out 3 solutions
then z^3+1= the 3 solutions and work out 3 more from there

[melissa.]

i can only get 7 answers =/

*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

[gta007]

OKay thanks alot xZero and melissa.
I'll let my friend know, and report back in if there is any further assistance required.

Thanks again.

[melissa.]

btw, are you sure there are 9 answers?

[gta007]

That's what I was told. She has the answers anyway.  :S

Once again, thanks for the assistance.

[melissa.]

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

[Sunny10]

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

[TrueTears]

Clearly has 9 solutions over by the fundamental theorem of algebra (http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra) [Although some may not be unique]



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.

[TrueTears]

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

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)

[melissa.]

hmm
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

[TrueTears]

You would say there is only 1 unique solution but 3 if we include each root's multiplicity.

Quote from the wikipedia page:

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.

[melissa.]

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.

[TrueTears]

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.

No worries, I just did so for you

Bahahaha, I saw AzureBlue's karma first and I was like O: until I saw this reply lulz