Page 1 of 2 12 LastLast
Results 1 to 15 of 19
Like Tree7Thanks

Math Help - Confusion with'i'

  1. #1
    Newbie
    Joined
    Mar 2014
    From
    Switzerland
    Posts
    2

    Confusion with'i'

    Ok hey guys,
    I have sort of confused myself with the following:
    \frac{1}{i}\cdot\sqrt{\frac{1}{i}} = -i\left(\frac{1}{i}\right)^{\frac{1}{2}}=-\left(\frac{i^2}{i}\right)^{\frac{1}{2}}=-\sqrt{i}
    \frac{1}{i}\cdot\sqrt{\frac{1}{i}} = \left(\frac{1}{i^3}\right)^{\frac{1}{2}} = \left(\frac{-i}{i^2}\right)^{1/2} = \sqrt{i}.

    So which one is correct?? And which operation is illegal/ill-defined?
    Thanks a lot for the help.
    Last edited by Looknan; March 1st 2014 at 04:33 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Confusion with'i'

    The symbol:
    $$\sqrt{i}$$

    is not "well-defined": the complex number i has TWO square roots, and there is no clear choice for making one "the positive" one.
    Thanks from Plato, Looknan and Shakarri
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Confusion with'i'

    They are both correct. i2 has two roots, +i and -i
    If you use the negative root in the first line you get:
    -(i2/i)1/2=-((-i)(-i)/i)1/2 = (i)1/2
    Thanks from Looknan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2014
    From
    Switzerland
    Posts
    2

    Re: Confusion with'i'

    Ok, I feared that's what's happening. Thanks for clearing that up!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,242
    Thanks
    1795

    Re: Confusion with'i'

    Notice Hartlw saying that " i^2 has two roots, i and -i" and Deveno saying "there is no clear choice for the positive one". They are both correct. Since the complex numbers do NOT form an "ordered" field, there is no single choice for "positive" and "negative" complex numbers.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Confusion with'i'

    Quote Originally Posted by Deveno View Post
    The symbol:
    $$\sqrt{i}$$

    is not "well-defined": the complex number i has TWO square roots, and there is no clear choice for making one "the positive" one.
    In the complex number system, c1/n is very well defined. That's why the complex number system was created.

    And anyhow, the square roots of i have nothing to do with OP. It's the square roos of i2 that explains the OP.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie Informalsports's Avatar
    Joined
    Feb 2014
    From
    Pakistan
    Posts
    12

    Re: Confusion with'i'

    Me also Confuse what are you doing now friends ?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Confusion with'i'

    Why does an incorrect and irrelevant post get three thanks, and when corrected (#6), a retard suddenly appears for the first time out of nowhere to upstage the correction?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,922
    Thanks
    1762
    Awards
    1

    Re: Confusion with'i'

    Quote Originally Posted by Looknan View Post
    Ok hey guys,
    I have sort of confused myself with the following:
    \frac{1}{i}\cdot\sqrt{\frac{1}{i}} = -i\left(\frac{1}{i}\right)^{\frac{1}{2}}=-\left(\frac{i^2}{i}\right)^{\frac{1}{2}}=-\sqrt{i}
    \frac{1}{i}\cdot\sqrt{\frac{1}{i}} = \left(\frac{1}{i^3}\right)^{\frac{1}{2}} = \left(\frac{-i}{i^2}\right)^{1/2} = \sqrt{i}.
    Have a look at this discussion.
    There is a growing trend away from using $\sqrt{t}$ unless $t\in\mathbb{R}$.

    Here is an early theorem in Marsden/Hoffman: If $z\in\mathbb{C}$ then $\exists w\in\mathbb{C}$ such that $w^2=z$ (notice that $-w$ also works).
    We would use $-i\left(\frac{1}{i}\right)^{\dfrac{1}{2}}$ or $-i\left(-i\right)^{\dfrac{1}{2}}$

    Because $\mathbb{C}$ is not ordered it is difficult to talk about a principal root.
    But some authors do it this way.
    If $z\in\mathbb{C}$ If $z^{\dfrac{1}{2}}=\sqrt{|z|}$$\exp\left(\dfrac{i \text{Arg}(z)}{2}\right)$

    If you use that convention then $(-i)^{\dfrac{1}{2}}=\exp\left(\dfrac{-i\pi}{4}\right)=$$\left(\dfrac{1}{\sqrt{2}}\right) +\left(\dfrac{-i}{\sqrt{2}}\right)$
    Thanks from Looknan
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Confusion with'i'

    Last post by Plato totally irrelevant, of course.

    In the complex number system, c1/n is very well defined. That's why the complex number system was created.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,922
    Thanks
    1762
    Awards
    1

    Re: Confusion with'i'

    Quote Originally Posted by Hartlw View Post
    Last post by Plato totally irrelevant, of course.
    In the complex number system, c1/n is very well defined. TRUE
    That's why the complex number system was created. Absolutely FALSE
    I understand that you think you are able to help with higher mathematics but you just know enough about the subject.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,922
    Thanks
    1762
    Awards
    1

    Re: Confusion with'i'

    Quote Originally Posted by Hartlw View Post
    Why does an incorrect and irrelevant post get three thanks,
    Maybe that are four people who actually know enough about complex numbers to comment.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,922
    Thanks
    1762
    Awards
    1

    Re: Confusion with'i'

    Quote Originally Posted by Looknan View Post
    Ok hey guys,
    I have sort of confused myself with the following:
    \frac{1}{i}\cdot\sqrt{\frac{1}{i}} = -i\left(\frac{1}{i}\right)^{\frac{1}{2}}=-\left(\frac{i^2}{i}\right)^{\frac{1}{2}}=-\sqrt{i}
    \frac{1}{i}\cdot\sqrt{\frac{1}{i}} = \left(\frac{1}{i^3}\right)^{\frac{1}{2}} = \left(\frac{-i}{i^2}\right)^{1/2} = \sqrt{i}.
    @looknan, Please study this link.
    As it shows that real operations may(rules) not obey apply to complex numbers.
    However, complex operation are compatible with real numbers.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Confusion with'i'

    Quote Originally Posted by Hartlw View Post
    Last post by Plato totally irrelevant, of course.

    In the complex number system, c1/n is very well defined. That's why the complex number system was created.
    In actual fact, there are "difficulties" with defining c1/n in the complex numbers. To see why, let's look at the analogous definition in the real numbers:

    c1/n = cx = elog(c)x for x = 1/n.

    The trouble is: "log" isn't definable on ALL the complex numbers because the analogue of ex isn't 1-1 on the complex numbers anymore (so it doesn't have a UNIQUE inverse: log).

    To get around this difficulty, one can "restrict" the domain of the complex exponential to a horizontal "swath" of height 2*pi (this is what the "Arg" function does, but there are actually "many" arg functions we could use, so we have to be CLEAR about "which one" we are using).

    The two complex numbers:

    $$\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$$
    $$\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$$

    BOTH square to i, and there is no clear-cut choice for calling either one THE square root of i.

    In a similar vein, there is no reason to say i OR -i is THE square root of -1. In fact, choosing EITHER ONE leads to a version of the complex numbers that we "can't tell apart algebraically". This is because:

    (z + w)* = z* + w*
    (zw)* = z*w*

    So the mapping z-->z* preserves all the algebraic properties of the complex numbers (it is a field automorphism). The choice of i as the "positive direction of the y-axis" is a completely arbitrary one.
    Thanks from romsek and Plato
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,922
    Thanks
    1762
    Awards
    1

    Re: Confusion with'i'

    Let $a,~b\in\mathbb{C},~~ a\ne 0$ then $a^b=e^{b\log(a)}$

    Here is an excellent quote from Jerrold E. Marsden: $a^b$ is singled valued if and only if $b\in\mathbb{Z}$; if $b\in\mathbb{Q}$ where $b=p/q$ in lowest terms then $a^b$ has exactly $q$ distinct values; if $b\in\mathbb{R}\setminus\mathbb{Q}\text{ or if Im}(b)\ne 0$ then $a^b$ has infinitely many values.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Confusion...
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 2nd 2013, 05:35 AM
  2. Max and Min confusion
    Posted in the Differential Geometry Forum
    Replies: 9
    Last Post: May 18th 2011, 09:44 AM
  3. Set confusion
    Posted in the Discrete Math Forum
    Replies: 17
    Last Post: January 28th 2010, 10:00 AM
  4. confusion
    Posted in the Statistics Forum
    Replies: 2
    Last Post: August 15th 2009, 03:17 AM
  5. a confusion
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 24th 2008, 02:59 AM

Search Tags


/mathhelpforum @mathhelpforum