Results 1 to 4 of 4
Like Tree2Thanks
  • 1 Post By Plato
  • 1 Post By Vlasev

Thread: Can 2 times the dot product of 2 complex numbers equal to this?

  1. #1
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    308
    Thanks
    16

    Can 2 times the dot product of 2 complex numbers equal to this?

    There's a section in Gibson's "Elementary Geometry of Differentiable Curves:An Undergraduate Introduction" book on page 6:

    "Example 1.4: Recall that the component of a vector $\displaystyle a$ in the direction of a unit vector $\displaystyle b$ is the vector $\displaystyle (a \cdot b)b$. (Example 1.1)
    It is useful to express this in complex notation. Note that for any vectors $\displaystyle a,b$ we have $\displaystyle 2(a \cdot b) = a \overline{b} + b \overline{a}$: in particular when
    b is a unit vector (i.e. $\displaystyle b \overline{b} = \left | b \right |^2 = 1$) we have $\displaystyle 2(a \cdot b)b = a + \overline{a} b^2$."


    What I don't understand is why and how is: $\displaystyle 2(a \cdot b) = a \overline{b} + b \overline{a}$?

    The author said(given above) and I quote "for any vectors" this is true.

    Can anyone kindly shed light on this and explain a bit why this is so?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,782
    Thanks
    2824
    Awards
    1

    Re: Can 2 times the dot product of 2 complex numbers equal to this?

    Quote Originally Posted by x3bnm View Post
    "Example 1.4: Recall that the component of a vector $\displaystyle a$ in the direction of a unit vector $\displaystyle b$ is the vector $\displaystyle (a \cdot b)b$. (Example 1.1)
    It is useful to express this in complex notation. Note that for any vectors $\displaystyle a,b$ we have $\displaystyle 2(a \cdot b) = a \overline{b} + b \overline{a}$: in particular when
    b is a unit vector (i.e. $\displaystyle b \overline{b} = \left | b \right |^2 = 1$) we have $\displaystyle 2(a \cdot b)b = a + \overline{a} b^2$."
    What I don't understand is why and how is: $\displaystyle 2(a \cdot b) = a \overline{b} + b \overline{a}$?
    Suppose that $\displaystyle a=x+iy~\&~b=u+iv$.
    You can show that $\displaystyle a\overline{b}+\overline{a}b=2(xu+yv)$.
    But as vectors $\displaystyle a\cdot b=xu+yv.$
    Thanks from x3bnm
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    17

    Re: Can 2 times the dot product of 2 complex numbers equal to this?

    We have

    $\displaystyle a\overline{b}+\overline{a}b = a\overline{b}+\overline{a\overline{b}} = 2Re(a\overline{b})$

    The real part is just

    $\displaystyle a_1b_1-a_2(-b_2) = a_1b_1+a_2b_2 = a\cdot b$
    Thanks from x3bnm
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    308
    Thanks
    16

    Re: Can 2 times the dot product of 2 complex numbers equal to this?

    Thanks a lot Plato and Vlasev. That makes sense now. Again Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Product of complex numbers in polar form
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Jul 19th 2011, 08:03 AM
  2. complex numbers - product of polar forms
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Apr 10th 2011, 01:13 PM
  3. Product of Complex Numbers proof
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Dec 8th 2010, 02:31 PM
  4. Product of complex numbers.
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Sep 12th 2010, 10:09 AM
  5. find the product and its conjugate-complex numbers
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Aug 11th 2009, 10:37 AM

Search Tags


/mathhelpforum @mathhelpforum