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

[dyn]

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

 

[fresh_42]

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.

 

[dyn]

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

 

[fresh_42]

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.

 

[dyn]

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

 

[fresh_42]

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: $z=re^{i\varphi}$.