Complex inequality

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[solved]Complex inequality

Homework Statement


You have the two inequalities, where k is a complex number;
[tex] |k+\sqrt{k^2-1}|<1[/tex]
and
[tex] |k-\sqrt{k^2-1}| <1[/tex]

Show that if ##|k|>1##, then the second inequality is fulfilled, while the first one is impossible for any value of k.

The Attempt at a Solution



Those absolute-value signs freak me out.. Can somebody show me what to do? I'm sure this should be really easy but my brain is totally burnt out right now
 
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Answers and Replies

  • #2
CompuChip
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Two observations:

|x| < 1 can also be written as -1 < x < 1.

|k| > 1 implies k² > 1 (because k² = (-k)²).

The former should help you to rewrite the problem without absolute value signs.

[EDIT] I just noticed you mentioned k is complex, which means they are not absolute value signs but modulus signs (i.e. if ##k = r e^{i \phi}## then ##|k| = r##. That probably makes my hints above useless, lemme think about it some more.
 
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  • #3
PeroK
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I'm not sure I believe this. The square root of a complex number in not uniquely defined. Unlike real numbers, there is no way to choose "the positive one".

So, if you take k = -2i, you get k^2-1 = -5. Then:

[itex]k \pm \sqrt{k^2 - 1} = -2i \pm \sqrt{5}i[/itex]

Which has modulus > or < 1 depending on which root you take.
 
  • #4
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oh crap I'm so sorry, I mixed it up. k is non-complex!! I needed help with that inequality as it came up when I was solving a real integral by method of residues, but after I transformed the integral into a complex one, I forgot that k was real (it was a constant in an ordinary integral: ##\int_0^\pi \frac{2 d \theta}{k-\cos( \theta)}##).

Sorry! I didn't mean wasting anybody's time. I think I should be able to solve it now after this insight.... Thanks for the help
 
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  • #5
CompuChip
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Sorry! I didn't mean wasting anybody's time. I think I should be able to solve it now after this insight.... Thanks for the help

You didn't. If we had a life we wouldn't be on this forum :-)

(At least speaking for myself, don't mean to offend anyone).
 

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