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Complex numbers

  1. Mar 27, 2013 #1

    I'm trying to figure out whether this expression √(a^2+2ab-2ac+b^2+2bc+c^2) can be complex or not while a>0, b>0 and c>0.

    My answer would be no but I'm not sure.
  2. jcsd
  3. Mar 27, 2013 #2
    What is the only way for


    to be complex?

    Applied to your case, you know that [itex]a>0,\,b>0,\,c>0[/itex], do you see what term would be the only way to obtain the condition you just found? And is that even possible in this case? Suggestion: once you found the term that can lead to what you seek, use a well known binomial formula.
  4. Mar 27, 2013 #3


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    Staff: Mentor

    Every real number is a complex number as well. If you are looking for complex numbers with non-zero imaginary part, see the reply above.
  5. Mar 27, 2013 #4

    x has to be negative ?

    The expression can be complex if 2ac>b^2+2bc+c^2+a^2+2ab ?

    I can't really see what the term should look like ?
  6. Mar 27, 2013 #5

    Your on the right track. In order to have an imaginary part - mfb reminded that each real number is also a complex number - the expression under the square root needs to be negative.

    You pointed [itex]2ac[/itex]. Do you see two other terms in the expression you could associate with it and use a binomial formula?
  7. Mar 27, 2013 #6
    well I see the equation a^2+c^2+2ac but I'm not sure why or how to use the binomial formula ?
  8. Mar 27, 2013 #7
    Well, that's [itex] a^2 - 2ac + c^2[/itex] first. That doesn't ring a bell?
  9. Mar 27, 2013 #8
    Not really :S I can't see how I can use a binomial formula ? Is it something like this (a-c)^2=-r,

    r[itex]\in[/itex]ℝ ?
  10. Mar 27, 2013 #9
    Well, that's it:

    [itex]a^2 - 2ac + c^2 = \left ( a-c\right)^2[/itex]

    Can this become negative if [itex] a,c \in \mathbb{R} [/itex]?
  11. Mar 27, 2013 #10

    Yes it can become negative.
    Last edited: Mar 27, 2013
  12. Mar 27, 2013 #11
    Can it? Really? Can [itex]x^2[/itex] be negative if [itex]x\in\mathbb{R}[/itex]?
  13. Mar 27, 2013 #12
    Oh no of course it can't !
  14. Mar 27, 2013 #13
    So, now that you have established that [itex]\left( a-c\right )^2[/itex] cannot be negative, what can you say about [itex]\left( a-c\right )^{2}+2ab+b^2+2bc[/itex]?
  15. Mar 27, 2013 #14
    since a>0, b>0, c>0 and [itex]\left( a-c\right )^2[/itex] > 0, the expression [itex]\left( a-c\right )^{2}+2ab+b^2+2bc[/itex] must be positive.
  16. Mar 27, 2013 #15
    And now it is time to conclude!

  17. Mar 27, 2013 #16

    Thank you jfgobin :)
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