Solving Complex Inequalities with Absolute Value Signs

[Nikitin]

[solved]Complex inequality

Homework Statement

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

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

 

[CompuChip]

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, let me think about it some more.

 

[PeroK]

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:

$k \pm \sqrt{k^2 - 1} = -2i \pm \sqrt{5}i$

Which has modulus > or < 1 depending on which root you take.

 

[Nikitin]

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

 

[CompuChip]

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).