# what means to be outside unit circle?

I am trying to study time series without a great math background and I came across the next problem:

When checking for stationarity I check the roots, and if they are not on the unit circle, then it confirms the ts is stationary.

My question might be pretty basic: when I get the roots, I am getting scalars, whereas in my mind to be inside/outside unit circle I need coordinates... How do I know if a scalar is inside/outside the unit circle?

And my last question is, I always read "roots are not on the unit circle".

"on" means "inside", right? and if so... Is it the same as saying "roots are outside the unit circle"?

• You are getting real scalars, then, which by definition have $0$ for their second coordinates. Thus, $x$ lies inside the unit circle if and only if $(x,0)$ does, which means $|x|\lt 1.$ "On" literally means on the circle, not its interior (which is a disk). Thus, $x$ is on the unit circle if and only if $x=\pm 1.$ For intuition, see the examples at stats.stackexchange.com/a/29129/919. – whuber Nov 26 '19 at 15:06

The roots in this case are roots of a polynomial, and they can be (and often are) complex numbers. That means they have coordinates, in this case called the real part and the imaginary part. As an example, the polynomial $$z^2+1$$ has roots, that are solutions of the equation $$z^2+1=0$$, equal to $$z_{1,2}= \pm \sqrt{-1}=\pm i$$ where $$i=\sqrt{-1}$$ is the imaginary unit.
In this example, the real and imaginary parts are \begin{alignat}{4} \Re i &=&0 &,& \quad \Im i &=& 1 \quad \text{and}\\ \Re{-i}&=&0 &,& \quad \Im{-i} &=&-1 \end{alignat} and with pairs (or coordinate) representation $$i=(0,1), -i=(0,-1)$$ and now you can show these two roots are on the unit circle.