complex numbers

Find the roots of the equation (z+1)^3 = 1, in the form x+iy.

gogogo

Do your own homework.

Christ, they'd ask a problem that easy this late in the term? Where are you studying, Mexico City Highschool Academy?

>>3
How did you know?

1 = e^(i*pi*k) (is that right?)

find z for k = 1,2,3 etc.

Use DeMoivre's, it saves a huge amount of time.

(x+1)^3 = (x+1)(x^2+2x+1) = x^3+2x^2+x+x^2+2x+1 = x^3+3x^2+3x+1

Problem: roots of (x^3+3x^2+3x = 0)

x(x^2+3x+3) = 0 (<--This makes zero one of the roots)

Roots of (x^2+3x+3):
[-3+/-sqrt(9-4(1)(3))]/2 (<--Quadratic Formula)
-1.5+i*sqrt(3/4) and -1.5-i*sqrt(3/4) (<--I solved it in my head to the form of (a+bi) like complex numbers are supposed to be written.)

There.  Turn that in and prove that you don't understand it.

>>7
Or you could take the third roots of 1, and subtract 1 from each.

>>6
Isn't that a little complex for this problem?

>>7
You did that on purpose, huh?  Sneaky.

>>8
I dunno what you're talking about, but I like it.

Since the assignment is probably past due, I wanna try.

Normally you would just find the roots of the terms inside the brackets, but the 1 complicates things.  it must first be subtracted.  You then do the trinomial expansion of the factored polynomial.  You can find the terms by using Pascals triangle and the form  a^n*b^0 + a^n-1*b^1 + ... + a^0*b^n.  (or just use the trinomial expansion theorem). 

Next you need to combine your terms.  The -1 will cancel at this point.  You should be able to factor, as 7 did, making one of your roots 0, and giving you a quadratic equation to solve.  Once you've solved for your solutions, if there is a negative square root, factor that out as i, and you have your answer.

Good job 7.  You da man.

>>8
>>9
this means:
(x+1)^3 ^1/3 = 1 ^1/3
(x+1) = 1
x = 0

obviously this is only 1/3 of the complete answer.

>>10
I c wut u did ther.  :)

>>10
No, it means:
(x+1)^3 = 1 -> x+1 is a third root of 1. Therefore x+1 = 1, (-1 + i*sqrt(3))/2, or (-1 - i*sqrt(3))/2. Therefore x = 0, (-3 + i*sqrt(3))/2, or (-3 - i*sqrt(3))/2.

>>12

3 roots, of which he found 1.  Therefore he had 1/3 of the answer.  It doesn't take a genius, but it helps.

>>9
What did I do that was sneaky?  I expanded the parentheses (which I should have done with Pascal's Triangle instead of the Distributive Property), subtracted 1 from both sides, and factored out x from the expression containing it.  That left me with x*[quadratic], which made 0 one root by default.  Then, I used the Quadratic Formula to find the two roots of the remaining quadratic, and came up with -1.5 +/- i*sqrt(3/4) for the other two roots.  There's nothing hard about the problem.

Algebra II, I think?  In any case, it was a very straightforward and easy problem.

>>12
You're almost right.  You failed to do something correctly.  I'm not quite sure what it is, based on your post, but the way I did it was through the quadratic formula, in which the -b term is also divided by 2*a.  You've got -3 where -3/2 or -1.5 should be.

Also, >>9
It's easier to expand and then subtract because once you've expanded, you've got a free-standing 1 as it's own term, and not as part of another term like (x+1)^3.  Subtracting after expansion means that the proper 1's drop out immediately when necessary.  All I'm saying is that you should reverse your first two steps, and you're good.  Sorry I failed to catch this in my last post.

>>14
You got the right answer.  very clever of you.  Almost...too clever.

Don't worry, was being smartass...or dumbass, either way lots of ass, so it's good, right?

next time you need roots. take your function of the form:

something^anything = anotherthing and make it

something^anything - anotherthing = 0

graph it using some fancy Ti - 83 against zero and look where they intersect.

>>17
lol@TI.  HP39G, please!  Also, why would you have to graph against zero?  Why not just look for intersection with the x-axis?

>>18 you can do the 'find interestion between the functions' thing with the extra 0 plotted. Also, please forgive me, I donno what kids use as calculators these days.

>>15
Look more carefully at my parentheses, the -3 is over 2.

Elaborating on >>5...


(z+1)^3 = 1 = e^(i*2*pi*k)
z+1 = (e^(i*2*pi*k))^(1/3) = e^(i*2*pi*k/3)

For the following, we are going around the complex plane looking for instances where e^(i*t) = 1. The first three instances will give us all the 3 solutions. When k=3, we have gone a full circle around the complex plane.

k=0: z+1 = e^(0/3) = 1
=> z = 0

k=1: z+1 = e^(i*2*pi/3) = cos(2/3 pi) + i sin(2/3 pi) = -1/2 + i sqrt(3)/2
=> z = -3/2 + i sqrt(3)/2

k=2: z+1 = e^(i*4*pi/3) = cos(4/3 pi) + i sin(4/3 pi) = -1/2 - i sqrt(3)/2
=> z = -3/2 + i sqrt(3)/2

>>20
I know what it is.  All complex numbers must be written in the form of a+bi.  Your answer is in the form (a+bi)/c, c being a constant, which is technically "wrong".  You'll lose points and credibility for writing something like that when it matters.

>>22
HA HA OH WOW

>>22
Oh sorry, I didn't realize I was talking to a fucking middle school student. Just disregard my answer and forget I said anything.

>>24
R-R-R-R-RAAAAPED!

>>22

I know what it is.  All complex numbers must be written in the form of R e^(i theta).  Your answer is in the form a+bi, a and b being constants, which is technically "wrong".  You'll lose points and credibility for writing something like that when it matters.

>>24,
I'm in college, dumbass.  And at Carnegie Mellon University, no less.  I know my math.  For a reference point of my mathematical knowledge, I'm taking 21-127, Concepts of Mathematics.  This course focuses on theorems, proofs, and thinking straight, with particular emphasis on accuracy of proofs.  This course is higher level than 21-122, a Calculus course, which most people in my class (Freshman) are taking.

Therefore, I know what I'm talking about.

>>26
Do you really not know what a complex number is, or are you just being a total fucking faggot?

In any case, >>24 and >>26, you both need to learn more math.  Start here:

http://en.wikipedia.org/wiki/Complex_number

>>27
you sound like a faggot who doesn't understand what he's talking about.  no professor who wansn't on an ego trip or a complete douchebag would take off points for writing something like (a+bi)/c or think any less of you, unless they specifically asked for a+bi form.  also, re^i(theta) notation for complex numbers and equations is regularly a shitload more useful, so youre either a troll or a dipshit.  you're probably one of those guys who was a nerd in highschool and thinks hes awesome now in his first semester of college because he got an A on the midterm, but who is mostly a social failure and a douchebag.  so, as a social non-failure, with more background in mathematics than you, i say the following: quit being a retarded bitch.

>>28
I'm guessing you didn't go to Central Catholic High School in Pittsburgh, Pennsylvania.  You never experienced Mr. J.D. Williamson's AP Calculus course.  Perfection per problem is mandatory, unless you really want to lose between 1/4 and 1/2 inclusive points on any given question.  I've never heard of this re^i(theta) form, but I took another at the article I linked, and I see that it is a Polar form (which I should have guessed).  Oops on my part.

But I am by no means a retarded bitch, and the "social failure" issue is one that I'm working on (Don't remind me that I'm on 4chan.  I know.).  I wasn't exactly a nerd in high school, and most of the people in my class at CMU are way smarter than me.  But please, tell me, what is your background in mathematics, as you claim to have more than me?

Also, I'm trained by GameFAQs to not be a troll or a dipshit.  No further comment on this point.

You should probably realize that since in a+bi, a and b can be any constants at all (much like the C in indefinite integration), and therefore, if a+bi is a quantity divided by a constant c (as in (a+bi)/c), one could and should just divide the c into a and b and write the complex number as a+bi.

>>29
>>24 here. I don't know what >>28's background in mathematics is, but I'm presently working on a PhD. >>27 is completely correct in saying that no professor (or TA, or department grader) would complain about (a+bi)/c unless it was explicitly stated that your answers should be in the form a+bi (or r*e^(it)). You had a shitty high school teacher; an experience most people will find familiar. That doesn't make your idiotic statement any less true. Do CMU a favor and quit tarnishing its reputation by pretending to know what you're talking about with regards to mathematics.

>>30
Oops, that should read ">>28 is completely correct..."

>>29

Oh, AP Calc? You have it so tough. PROTIP: Learning partial derivatives doesn't make you any smarter than anyone else taking single variable calculus.

Also, catholics suck, ergo you suck. CFQD.

>>30
I do know what I'm talking about.  I make a point of not discussing subjects I'm not familiar with.  The standard for expression of complex numbers is a+bi (or, apparently, re^i*theta)).  This makes (a+bi)/c technically incorrect, because a/c and b/c are also constants.  Convert these rational numbers to decimal form, and you get a new (yet equivalent) a and b for your complex number, which can now be written in a+bi.  Of course, (a/c + i*b/c) is also acceptable.  This is what I was taught.  If (a+bi)/c is now a conventional and valid form of complex numbers, please tell me and link me to a source.

>>32
Fuck you.
(a) I -had- it so tough.  Note that I'm no longer in high school.
(b) Catholics do not suck.  And what the fuck is "CFQD"?

>>33
"CFQD" = can't fucking quit dickrubbing

you fucking dickrubber rubbing your dick on the furnace you bitch mother

>>34, fuck you.

OOOH, BITCHFIGHT

>>33
>(b) Catholics do not suck.

I consider that to be highly unlikely.

Yeah, I gave that Catholicism shit up at... what... 12?

this thread is awesome.
entertain me more.

this thread has stopped being about helping a kid with his homework and is now about epeens, post some fucking epeens!