Complex numbers

  • Thread starter Tala.S
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  • #1
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Hi

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.
 

Answers and Replies

  • #2
90
2
What is the only way for

[itex]\sqrt{x}[/itex]​

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.
 
  • #3
35,437
11,856
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.
 
  • #4
43
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What is the only way for

[itex]\sqrt{x}[/itex]​

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.


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 ?
 
  • #5
90
2
Tala,

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?
 
  • #6
43
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Tala,

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?

well I see the equation a^2+c^2+2ac but I'm not sure why or how to use the binomial formula ?
 
  • #7
90
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Well, that's [itex] a^2 - 2ac + c^2[/itex] first. That doesn't ring a bell?
 
  • #8
43
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Well, that's [itex] a^2 - 2ac + c^2[/itex] first. That doesn't ring a bell?

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 ?
 
  • #9
90
2
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]?
 
  • #10
43
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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]?


Yes it can become negative.
 
Last edited:
  • #11
90
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Can it? Really? Can [itex]x^2[/itex] be negative if [itex]x\in\mathbb{R}[/itex]?
 
  • #12
43
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Can it? Really? Can [itex]x^2[/itex] be negative if [itex]x\in\mathbb{R}[/itex]?

Oh no of course it can't !
 
  • #13
90
2
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]?
 
  • #14
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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]?

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.
 
  • #15
90
2
And now it is time to conclude!

J.
 
  • #16
43
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Yes.

Thank you jfgobin :)
 

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