Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

We usually represent sqrt(-1) as i

  1. Jun 30, 2008 #1
    we usually represent sqrt(-1) as i , but only a quadratic eqn or a eqn of high power would have led to represent the value as i.so for a quadratic eqn we obtain two values ,so why dont we have the two values as +i and -i.

    statement 1:
    X^2=4 so , X=+2 and -2
    Statement 2:
    X^2= -1 so X=+i and -i

    is statement 2 possible or my question is wrong??

    Have i asked a stupid q so that i should brush up my basics??
     
  2. jcsd
  3. Jun 30, 2008 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Re: iota

    i and -i are both solutions to x^2=-1. And it i isn't an iota; iotas don't have the dot on them.
     
  4. Jun 30, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: iota

    In fact, in the complex number system, any equation of the form xn= a, where a can be any complex number, has n distinct solutions.
     
  5. Jul 1, 2008 #4
    Re: iota

    So if i use -i instead of +i does it make any change ?
     
  6. Jul 1, 2008 #5

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Re: iota

    "Use"? How?

    There are 2 roots, they are indistinguishable. If we label one i, the other is -i. But there is no way to say which is which.
     
  7. Jul 1, 2008 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: iota

    This is going to get much more technical than the original poster will like! matt grime's point that "they are indistinguishable" is exactly why we CANNOT simply define "i" by "i2= -1" or as "[itex]\sqrt{-1}[/itex]". What can be done is to define the complex number system as a set of objects of the form (a, b) where a and b can be any two real numbers, and define addition and multiplication by (a,b)+ (c,d)= (a+ c,b+ d) and (a,b)(c,d)= (ac-bd, ad+ bc). If we identify with complex numbers of the form (a, 0), imaginary numbers with complex numbers of the form (0,, b) and define "i" as (0, 1), then we have (a, b)= (a,0)(1,0)+ (b,0)(0,1)= a+ bi. It then follows that i2= (0,1)(0,1)= (0(0)-1(1),1(0)+ 0(1))= (-1, 0)= -1. It is also true that (0,-1)(0,-1)= (0(0)- (-1)(-1), 0(-1)+ 0(-1))= (-1, 0)= -1 but we have specifically defined i to be (0, 1), not (0, -1).
     
  8. Jul 1, 2008 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: iota

    Use how and what kind of change? (-i)2= (-1)2(i)2= (1)(-1)= -1 = i2 but 1+ i is certainly different from 1- i. It makes the same change as if you use -1 instead of +1.
     
  9. Jul 1, 2008 #8
    Re: iota

    Thanks to all ur replies



    now i have another question .what would be the value of ii
     
  10. Jul 1, 2008 #9

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Re: iota

    It has infinitely many 'values', since raising to the i'th power is not single-valued. To make it single valued one chooses a so called branch.

    Note that a^b is defined to be exp(log(a^b)=exp(b log a) in general, so it is the same as defining log(i). Write i in polar coordinates to do this, and choose Arg (i.e. pi/2).
     
  11. Jul 1, 2008 #10
    Re: iota

    Good question, one answer is ii = 0.20788...
     
  12. Jul 1, 2008 #11

    nicksauce

    User Avatar
    Science Advisor
    Homework Helper

    Re: iota

    To be exact, i^i = e^(-pi/2).
     
  13. Jul 1, 2008 #12

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: iota

    To be exact, that is one of the infinitely many values of [tex]e^{-pi/2}[/tex].
     
  14. Jul 1, 2008 #13

    nicksauce

    User Avatar
    Science Advisor
    Homework Helper

    Re: iota

    True
     
  15. Jul 3, 2008 #14
    Re: iota

    [tex]e^{-\frac{\pi}{2}}[/tex] has multiple values? Since when?
     
  16. Jul 3, 2008 #15

    nicksauce

    User Avatar
    Science Advisor
    Homework Helper

    Re: iota

    i^i has multiple values. That one is one of them.
     
  17. Jul 3, 2008 #16

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: iota

    I was about to retort that no one had said any such thing- then noticed my typo.

    I meant to say that [itex]e^{-\pi/2}[/itex] was one of the infinitely many values of [itex]i^i[/itex].
     
  18. Jul 3, 2008 #17
    Re: iota

    I see.

    I suppose it depends on how you define [tex]i^i[/tex]. My book defined it as [tex]e^{Log(i)*i}[/tex] which was in turn defined completely unambiguously...

    And by "my book" I mean "the only book I happen to have looked at that talks about complex exponential functions"
     
  19. Jul 3, 2008 #18

    nicksauce

    User Avatar
    Science Advisor
    Homework Helper

    Re: iota

    I suppose that this is what one might call the principal value of i^i, by taking Log(i) = log(r) + i*theta, where theta is in the range[0,2pi). But you can also have other values by modifiying Log(i) by any multiple of i*2pi.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?