How to know if a complex root is inside the unit circle

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  • #1
dyn
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Hi.
I have been trying to calculate the real definite integral with limits 2π and 0 of ## 1/(k+sin2θ) ##
To avoid the denominator becoming zero I know this means |k|> 1
Making the substitution ##z= e^{iθ}## eventually ends up giving me a quadratic equation in ##z^2## with 2 pairs of roots given by ##z^2 = i (+\sqrt{k^2-1} - k ) ##
and ## z^2 = i (-\sqrt{k^2-1} -k ) ##
The solution then states that"clearly the 1st two poles lie inside the unit circle and the 2nd two outside". This seems reasonable but how do I know for a fact that it is true ?
Thanks
 

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  • #2
fresh_42
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The second is easy. ##|i(-\sqrt{k^2-1}-k)|=|\sqrt{k^2-1}+k|>|k|>1##. The first should be obtainable by multiplying the two.
 
  • #3
dyn
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Thank you. Can I just check that I understand the rest ?
The quadratic equation in ##z^2## is ## z^4 + 2ikz^2 - 1 = 0 ##
The product of the 2 roots is αβ = -1 ⇒ |α|. |β| = 1
So knowing the modulus of one root is greater than 1 , means the modulus of the other root is less than 1

So that gives me , for one root ## |z^2| < 1 ## which also means for that root |z| < 1.
Have I got all that correct ?
Thanks
 
  • #4
fresh_42
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Thank you. Can I just check that I understand the rest ?
The quadratic equation in ##z^2## is ## z^4 + 2ikz^2 - 1 = 0 ##
The product of the 2 roots is αβ = -1 ⇒ |α|. |β| = 1
So knowing the modulus of one root is greater than 1 , means the modulus of the other root is less than 1

So that gives me , for one root ## |z^2| < 1 ## which also means for that root |z| < 1.
Have I got all that correct ?
Thanks
That was the idea, but I cannot see what you have done. ##z## is confusing here, as you use the same letter for different numbers. You haven't mentioned the quadratic equation up to now, only the solutions. But if ## z^4 + 2ikz^2 - 1 = 0 ## is your equation, I get ##z_{1,2,3,4} = \pm \sqrt{-ik \pm \sqrt{1-k^2}}##. I see what you have done to write it as you did. However, one has to be careful with complex numbers. Not every rule which is valid in ##\mathbb{R}## is also valid in ##\mathbb{C}##.
See: https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

I would therefore check the following equations, just to be sure:
\begin{align*}
&z^4+2ikz^2-1=\\
&=\left( z- \sqrt{-ik -\sqrt{1-k^2}} \right)\left(- \sqrt{-ik + \sqrt{1-k^2}} \right)\left( \sqrt{-ik - \sqrt{1-k^2}} \right)\left( \sqrt{-ik + \sqrt{1-k^2}} \right)\\
&=\left( z^2- i\left( \sqrt{k^2-1}-k\right) \right)\left(z^2-i\left(-\sqrt{k^2-1}-k\right) \right)
\end{align*}

Set ##\alpha=i\left( \sqrt{k^2-1}-k \right)\, , \,\beta=i\left( -\sqrt{k^2-1}-k \right)## and then we have ##\alpha \cdot \beta = - \left( -\sqrt{k^2-1}^2 +k^2 \right)= -1## and then what you wrote.
 
  • #5
dyn
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However, one has to be careful with complex numbers. Not every rule which is valid in ##\mathbb{R}## is also valid in ##\mathbb{C}##.
See: https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/
Do you think I maybe applied a rule valid in ℝ which is not valid in ℂ ? To get the sum of the products I used the fact that for a quadratic equation ## ax^2 +bx + c = 0 ## with 2 roots α and β then αβ = c/a which in the above case is -1 (and I have seen this same result applied to complex quadratic equations)
I presume it is just a coincidence that the product comes to +1 or -1 as this helps make it easier to find out if the roots lie inside the unit circle and if the product is not +1 or -1 it just makes it harder to figure this out
 
  • #6
fresh_42
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I haven't checked, probably not, but that's why I said "make the control calculation" or make sure you can extract ##i=\sqrt{-1}## from the root. The best way to check such things is the representation in polar coordinates: ##z=re^{i\varphi}##.
 

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