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

In summary, the real definite integral with limits 2π and 0 of ## 1/(k+sin2θ) ## can be simplified by using the substitution ##z= e^{iθ}## and solving for ##z^2##. This gives two roots α and β which lie inside and outside the unit circle, respectively. If the product of the two roots is not +1 or -1, this indicates that the roots lie inside or outside the unit circle, respectively.
  • #1
dyn
773
61
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
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.
 
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Likes dyn
  • #3
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
dyn said:
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
fresh_42 said:
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
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}##.
 

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

1. How do I determine if a complex root is inside the unit circle?

The easiest way to determine if a complex root is inside the unit circle is to plot it on a complex plane. If the root lies within the unit circle (a circle with a radius of 1 centered at the origin), then it is inside the unit circle.

2. Can I use any method other than plotting to determine if a complex root is inside the unit circle?

Yes, there are other methods you can use to determine if a complex root is inside the unit circle. One method is to calculate the magnitude of the root (the distance from the origin) and compare it to 1. If the magnitude is less than 1, then the root is inside the unit circle.

3. What is the significance of a complex root being inside the unit circle?

A complex root being inside the unit circle means that it has a magnitude less than 1, which indicates that it is a stable root. This is important in various fields of science, such as control systems and signal processing, as it indicates that the system will converge to a stable solution.

4. Can a complex root be both inside and outside the unit circle?

No, a complex root can only be either inside or outside the unit circle. If the root lies exactly on the unit circle, it is considered to be outside the unit circle.

5. How does the location of a complex root on the unit circle affect its behavior?

The location of a complex root on the unit circle can affect its behavior in different ways. If the root is on the unit circle, it is considered to be marginally stable, meaning that it can oscillate but will not diverge. If the root is inside the unit circle, it is stable and will converge to a solution. If the root is outside the unit circle, it is unstable and will diverge.

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