1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    2017 Award

    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 :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook