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  • File : 1269895769.jpg-(69 KB, 469x428, trollface..jpg)
    69 KB Anonymous 03/29/10(Mon)16:49 No.609268  
    tell me /sci/, what's:

    (i)
    >> Anonymous 03/29/10(Mon)16:52 No.609283
    inb4/sci/ is trolled again.

    But I would seriously like an answer to this and why.
    >> Anonymous 03/29/10(Mon)16:52 No.609286
    4th root of -1
    >> Anonymous 03/29/10(Mon)16:55 No.609301
    >>609286
    What's that in the form of a+bi?
    >> Anonymous 03/29/10(Mon)16:57 No.609310
    (i2)2
    >> Anonymous 03/29/10(Mon)16:58 No.609321
    +-(1+i)/sqrt(2)
    >> Anonymous 03/29/10(Mon)16:59 No.609325
    (-1±i)/SQR(2)
    >> Anonymous 03/29/10(Mon)17:01 No.609334
    i^(1/2) which becomes

    e^(i*pi/2)*(1/2) which becomes

    e^(i*(pi/4)), there are 3 other possible answers, maybe someone has already posted them..
    >> Anonymous 03/29/10(Mon)17:01 No.609335
    That wasn't very lulzy.
    >> Anonymous 03/29/10(Mon)17:04 No.609344
    >>609334

    other solutions are -e^(i*(pi/4)) and i*e^(i*(3pi/4)) and -i*e^(i*(3pi/4))
    >> Anonymous 03/29/10(Mon)17:04 No.609347
    >Report submitted! This window will close in 5 seconds...
    >> Anonymous 03/29/10(Mon)17:06 No.609357
    >>609347
    Despite being a troll thread, this is one of the few threads that actually has to do with science or math that I've seen in a while.
    >> Anonymous 03/29/10(Mon)17:06 No.609358
    >>609347
    No need to run. It's just math.
    >> Anonymous 03/29/10(Mon)17:06 No.609360
    >>609045

    wut?
    >> Anonymous 03/29/10(Mon)17:07 No.609362
    i^.5
    >> Josef !!cvJKiVKmWz2 03/29/10(Mon)17:08 No.609371
    >>609334
    It's a square root, thus there are only 2 solutions (see fundamental theorem of algebra).
    >> Anonymous 03/29/10(Mon)17:09 No.609374
    >>609371

    it's a square root of a square root

    now use your precious fundemental theorem of algebra again
    >> Josef !!cvJKiVKmWz2 03/29/10(Mon)17:12 No.609398
    >>609374
    i isn't the unique square root of any number. sqrt(-1) is nice to introduce i to highschool kids, but that's about it.
    >> Anonymous 03/29/10(Mon)17:14 No.609405
    >>609374
    so? 3 is a square root of 9. that doesn't mean that there's only one square root of 3.
    >> Anonymous 03/29/10(Mon)17:15 No.609410
    >>609398

    so what you're saying is.. x^4+1=0 has 2 solutions?
    >> Anonymous 03/29/10(Mon)17:19 No.609431
    >>609410

    It has 4 solutions
    >> Anonymous 03/29/10(Mon)17:20 No.609442
    So why does sqrt(i) have only 2 solutions, genuine question here.
    >> Anonymous 03/29/10(Mon)17:20 No.609448
    >>609380
    >thinks getting responses means trolling
    lol
    >> Anonymous 03/29/10(Mon)17:21 No.609451
    -1
    >> Josef !!cvJKiVKmWz2 03/29/10(Mon)17:21 No.609454
    >>609442
    "Square" stands for 2. 2 solutions.
    z4=1 uses a power of 4. Guess how many solutions.
    >> Anonymous 03/29/10(Mon)17:23 No.609463
    >>609442
    because c^z, for some complex number c and integer z, has exactly |z| solutions, some real, some complex
    >> Anonymous 03/29/10(Mon)17:24 No.609466
    >>609454

    so what you are saying is sqrt(sqrt(-1)) is not the same as sqrt(i)?

    in other words, i and sqrt(-1) are different?
    >> Anonymous 03/29/10(Mon)17:25 No.609473
    OK you fucking morons. You fools obviously know dick about any of this. Heard of this much?

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

    Any non-constant polynomial of a single variable over the complex field has as many roots as its degree.

    So the polynomial relation x^2+1=0 has TWO solutions. Similarly, x^n + 1 = 0 has n solutions. x^2 + i = 0 has TWO solutions, et cetera. Jesus Christ.
    >> Anonymous 03/29/10(Mon)17:28 No.609486
    >>609473

    nobody cares about your short temper for people who havent wikipedia'd as much stuff as you, if you look carefully people are actually asking questions (you know, that thing you do as a process of learning?), so either help them or go away.

    Thank you.
    >> Josef !!cvJKiVKmWz2 03/29/10(Mon)17:30 No.609503
    >>609466
    In function theory, you usually don't ditch one solution of a root like you do when using reals. Usually, all possible solutions are considered.
    Actually, I'm not sure whether the complex root has a precise definition at all - as far as I know, writing a=sqrt(z) is just a sloppy version of saying "let z be the set of solutions of a^2=z".
    >> Anonymous 03/29/10(Mon)17:30 No.609505
    >>609463
    >has |z| solutions

    wtf
    >> Anonymous 03/29/10(Mon)17:30 No.609507
    >>609486

    Bawwwww someone got their feelings hurt on 4chan.
    >> Anonymous 03/29/10(Mon)17:31 No.609512
    >>609486
    >implying you don't learn the statement of FTA in like 8th grade

    loludumb
    >> Josef !!cvJKiVKmWz2 03/29/10(Mon)17:31 No.609513
    >>609503
    typo; should've been "let sqrt(z) be the set of solutions of a^2=z".
    >> Anonymous 03/29/10(Mon)17:33 No.609522
    >trollface
    >checkmark (i)
    >+20 replies

    the samefag is strong in this thread
    >> Anonymous 03/29/10(Mon)17:33 No.609524
    >>609442
    Write out all 4 solutions to x^4 = -1. Two of them will square to i, and two will square to -i. There's your answer.
    >> Anonymous 03/29/10(Mon)17:35 No.609534
    >>609505has |z| solutions

    eg. x^-4 has 4 solutions
    >> Anonymous 03/29/10(Mon)17:36 No.609542
    >>609534

    ... do you even know wtf |z| means?
    >> Anonymous 03/29/10(Mon)17:39 No.609562
    >>609512
    >>609507
    >>609473

    Samefag, you've been asked once already, help out or fuck off back to /b/
    >> Anonymous 03/29/10(Mon)17:40 No.609575
    Well, since this is a "troll" thread, can I ask something itt completely unrelated to math or science that I've wanted to know for a while? I was just wondering how to post a message here with red text and how to make one of those spoiler boxes (the ones where you have to hover your mouse over it to read what's inside). I've seen it done here before and was just curious as to how it was done. Could someone please help. I won't do it here of course; perhaps I'll test it in /b/.

    btw I already know how to triforce; so if either of the things I inquired about require unicode, I've had previous experience. *shrug*
    >> Anonymous 03/29/10(Mon)17:42 No.609583
    >>609562

    I'm the only one who gave a concrete answer. I still don't know what you are not understanding.
    >> Anonymous 03/29/10(Mon)17:46 No.609608
    >>609583

    Posting a link from wikipedia about the fundemental theorem of algebra does not answer the question

    >>609466


    The only helpful people in here were Josef and some other non-nob head anons.
    >> Anonymous 03/29/10(Mon)17:52 No.609642
    >>609608

    I posted the Wikipedia page in lieu of actually writing out the statement of the theorem. This conversation seems to be revolving around the confusion over how many nth-roots there are of a certain complex number. The FTA answers this question precisely: there are n complex roots of any complex number. Therefore, the troll's original question can be stated differently: what are the solutions of x^{2} = i over the complex field? The answers are exp{i*\pi/4} and exp{i\5pi/4}.
    >> Anonymous 03/29/10(Mon)17:54 No.609655
    >>609608

    Since we have found two roots of the polynomial relation x^2 = i, the FTA guarantees that these two are unique.
    >> Anonymous 03/29/10(Mon)17:59 No.609677
    just visualize it on the complex plane. It would have 2 solutions, the first at 45 degrees and the second at 225 degrees. In other words it would be something like.

    cos(45) + i sin (45)
    and
    cos(225) + i sin(225)

    Alternatively you can write it in polar coordinates like this.

    e^((i)(pi/2))
    and
    e^((i)(5pi/4))

    It can also be represented in matrix coordinate form if you want to.
    >> Anonymous 03/29/10(Mon)18:00 No.609690
    >>609642
    Bawwwww someone got their feelings hurt on 4chan.
    >> Anonymous 03/29/10(Mon)18:02 No.609706
         File1269900176.jpg-(17 KB, 268x312, succesful troll.jpg)
    17 KB
    >>609268


    godamn it 40+ replies.



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