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complex numbers

Name: Anonymous 2007-10-31 19:26

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

gogogo

Name: Anonymous 2007-10-31 19:48

Do your own homework.

Name: Anonymous 2007-10-31 21:44

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

Name: Anonymous 2007-11-01 7:58

>>3
How did you know?

Name: Anonymous 2007-11-01 15:51

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

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

Name: Anonymous 2007-11-01 17:37

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

Name: anon 2007-11-01 21:41

(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.

Name: Anonymous 2007-11-01 22:12

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

Name: Anonymous 2007-11-01 23:51

>>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.

Name: Anonymous 2007-11-02 1:29

>>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.

Name: Anonymous 2007-11-02 1:43

>>10
I c wut u did ther.  :)

Name: Anonymous 2007-11-02 1:46

>>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.

Name: Anonymous 2007-11-02 1:50

>>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.

Name: anon 2007-11-02 1:50

>>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.

Name: anon 2007-11-02 1:57

>>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.

Name: Anonymous 2007-11-02 1:59

>>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?

Name: Anonymous 2007-11-02 2:25

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.

Name: anon 2007-11-02 11:14

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

Name: Anonymous 2007-11-02 20:38

>>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.

Name: Anonymous 2007-11-02 22:47

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

Name: Anonymous 2007-11-03 12:38

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

Name: anon 2007-11-04 16:33

>>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.

Name: Anonymous 2007-11-05 11:13

>>22
HA HA OH WOW

Name: Anonymous 2007-11-05 14:27

>>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.

Name: Anonymous 2007-11-05 15:36

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

Name: Anonymous 2007-11-05 23:45

>>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.

Name: anon 2007-11-06 21:29

>>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

Name: Anonymous 2007-11-06 21:38

>>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.

Name: anon 2007-11-06 21:56

>>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.

Name: Anonymous 2007-11-06 22:15

>>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.

Name: Anonymous 2007-11-06 22:16

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

Name: Anonymous 2007-11-06 22:17

>>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.

Name: anon 2007-11-06 23:43

>>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"?

Name: Anonymous 2007-11-07 0:24

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

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

Name: Anonymous 2007-11-07 1:10

>>34, fuck you.

Name: Anonymous 2007-11-07 1:21

OOOH, BITCHFIGHT

Name: Anonymous 2007-11-07 4:25

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

I consider that to be highly unlikely.

Name: Anonymous 2007-11-07 4:43

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

Name: Anonymous 2007-11-07 7:58

this thread is awesome.
entertain me more.

Name: Anonymous 2007-11-07 11:42

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

Name: Anonymous 2007-11-07 16:30

>>33
Simple question here: If you asked for a real number, and I responded "5/2", would you complain?

(Hint: it's one of those damned-if-you-do-damned-if-you-don't situations.)

Name: anon 2007-11-07 17:19

>>41
Of course I wouldn't complain.  "5/2" is a perfectly acceptible way of writing a real number.

By the way, how is it "damned if I do?"  I understand "damned-if-I-don't" because it's totally acceptible.  How does the first part play out?

Name: Anonymous 2007-11-07 18:15

>>33

Catholics DO suck. You'd have to be a true IDORT to believe that shit.

Name: Anonymous 2007-11-07 18:45

>>41
this is more analogous to being asked to give an integer and responding 42/7.

people might wonder why you didnt reduce it and format it the standard way, but no one really cares.

Name: Anonymous 2007-11-07 23:59

>>42
Because it's inconsistent to accept a ratio of real numbers as a real number when you obviously will not accept a ratio of complex numbers (indeed, a complex number divided by a real number, which is even nicer) as a complex number. Either you are incapable of going from (a+bi)/c -> a/c + i*b/c, or you are attempting to force your slow-witted high school teacher's idiotic grading policies on people who know better. In the former case, you're a fucking moron and you should gb2/middleschool/. In the latter case, you're a different kind of fucking moron, and you should kill yourself.

>>44
I would say that 5/2 (as a real number) and (a+bi)/c (as a complex number) are roughly equivalent concepts. Since complex numbers and real numbers both are fields, there is no worry about the division operation (provided c is nonzero). In the integers I'd be hesitant to give an answer in the form a/b even assuming a was a multiple of b, as division is not defined for all pairs of integers. I'm not saying 42/7 would be unacceptable, just that it would be considerably more odd than (a+bi)/c.

Name: Anonymous 2007-11-12 3:04

>>45
(a+bi)/c is a ratio of complex-to-real.  real-to-real is okay.  complex-to-real takes from two different fields.  A field only has closure for an operation if you can get back to the same field through the operation.  as complex and real are different fields, division isn't necessarily closed, mirite?

that's why (a+bi)/c is incorrect and a/c+i*b/c is correct.  the latter only has real-to-real ratios (one of them being multiplied by i), while the former has a complex over real which, like i said, takes from two different fields.

Name: Anonymous 2007-11-12 3:08

>>46
reals are in complex, complex to real is complex to complex.  now its a ratio of things from the same field.

Name: Anonymous 2007-11-12 4:25

>>46
So you read the word "field" somewhere on Wikipedia and decided to pull that out of your ass, did you?

Name: Anonymous 2007-11-12 4:29

>>48

LOL, that's true for most of replies in /sci/

Name: Anonymous 2007-11-12 4:56

>>46
How about Reel-to-Reel?  I Like to Move It Move It, I Like to Move It Move It, I Like to Move It Move It, bomp bomp bomp bum bum bomp bomp bomp bomp I Like to Move It Move It

Name: Anonymous 2007-11-12 11:04

Stop.



Hammertime.

Name: Anonymous 2007-11-12 11:15

>>48
no, but i've taken classes in abstract algebra, group theory, and two linear algebra courses.
are you suggesting that a real number is not a complex number?  >>46 was insinuating that a ratio of complex to real numbers may not be meaningful as a complex number because the numbers came from different fields.  if you're not a retard, you probably already know that the reals are a subset of the complex numbers, so any real number is a complex number.  so, while he would have a point if we were talking about whether or not the ratio made sense in the reals; we weren't.

Name: Anonymous 2007-11-12 11:24

>>52
I'm somewhat lost here. You seem to be telling >>48 that you know what you're talking about while simultaneously explaining why >>46 was wrong, despite the fact that >>48 was insulting >>46.

Name: Anonymous 2007-11-12 13:10

>>53
that's because i can't read and i thought >>48 was directed at >>47

woopzlol.

Name: Anonymous 2007-11-12 15:27

>>47
Then reduce for me the following:

(4+6i)/(8+10i)

Surely, this can be put into some kind of a+bi form, without remaining a quotient, right?

Name: Anonymous 2007-11-12 18:07

>>55
(23/41) + (2/41)i

Name: Anonymous 2007-11-12 19:12

>>55
relatedly
(a+bi)/(c+di) = [(ac+bd)/(cc+dd)] + [(bc-ad)/(cc+dd)]*i

this is defined as long as at least one of c or d are nonzero.  i.e. as long as the modulus of c+di is nonzero.

Name: Anonymous 2007-11-13 3:54

whats the fuckin difference between (a + bi)/c and a/c + bi/c ... the first is even better if you ask me

Name: Anonymous 2007-11-18 4:37

>>56
is (23+2i)/41 also an acceptible way to write that answer based only on this thread?

also, >>57,
is it also accpetible to write:
(a+bi)/(c+di) = [(ac+bd)+(bc-ad)i]/(cc+dd)

again based only on this thread?

Name: Anonymous 2007-11-18 11:26

>>59
i'm both 56 and 57.  but yeah, those are also an acceptable way to write both.  i don't know what 'based only on this thread' has to do with it.  if you have a working knowledge of fractions (ie, you didnt fail algebra), you should see that they are equivalent.
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