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Length of a complex vector

  1. Oct 22, 2012 #1
    Hello,

    I was going through a article which states as follows:

    If we had 2 real vectors like u & v, we would have calculated the length as
    √u^2+v^2

    but in case of a complex vector v = (1 + 2i)ˆi + (3 − 4i)

    we calculate

    as v.v= (1 + 2i)2 + (3 − 4i)2 = −10 − 20i

    My question is why do we multiply v.v, why not using the sqrt.formula?

    Thanks,

    -- Shounak
     
  2. jcsd
  3. Oct 23, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This was originally posted in "physics" and deleted because there are no "physics" questions in it. I have "undeleted" and moved to "mathematics" because there are so many errors that need to be addressed.

    You have misread. That is NOT "the length of two real vectors"- in fact, two real vectors do NOT have a single "length". It is, rather, the length of a single two-dimensional vector ui+ vj.

    The only correct answer is that we don't. You are mistaken- we do NOT find the length of compex vectors like that. For real numbers, we can define the "absolute value" as [itex]|x|= \sqrt{x^2}[/itex], a positive real number. For complex numbers that would not be a positive real number so we define, instead, [itex]|x|= \sqrt{xx^*}[/itex] where [itex]x^*[/itex] is the "complex conjugate"- the complex conjugate of the complex number a+ bi is a- bi so that [itex]|a+ bi|= \sqrt{(a+ bi)(a- bi)}= \sqrt{a^2+ bi- bi+ b^2}= \sqrt{a^2+ b^2}[/itex].

    For vectors with complex coefficients, ae1+ be2+ ce^3 where a and b are complex (I have used "e1", "e2", and "e3" as basis vectors rather than "i", "j", and "k" so as not to confuse the basis vector "i" with the imaginary unit, "i"), the length is the square root of the sums of the products of each component and its complex conjugate: [itex]\sqrt{aa^*+ bb^*+ cc*}[/itex]. If a, b, and c happen to be real numbers, then their complex conjugates are the same as a, b, and c themselves and the formulas reduces to the previous [itex]\sqrt{a^2+ b^2+ c^2}[/itex].

    In particular, the example you give, v = (1 + 2i)ˆi + (3 − 4i)^j has length [itex]\sqrt{(1+2i)(1- 2i)+ (3- 4i)(3+ 4i)}= \sqrt{1+ 4+ 9+ 16}= \sqrt{30}[/itex], a positive real number, NOT what you have.

    I recommend you go back and reread that article. Pretty much nothing you say you read in it is correct.

    I recomm
     
  4. Oct 23, 2012 #3
    Thanks a lot. Actually I was unable to get over the term for complex conjugate.

    I was confused with 1+2i^i means what? Raised to the power i or.......?

    Anyway, your answer was descriptive and it cleared my doubt.

    I didn't knew about how to calculate the length of a complex vector.

    Thanks,

    -- Shounak
     
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