I am considering the roots of the polynomial $$P(z)=z^2-z+\epsilon\lambda=0$$ with $\epsilon$ a small positive constant and $\lambda\in\mathbb{C}$ with $Re(\lambda)>0$. I am parrticularly interested to find conditions on $\epsilon$ (specifically upper bounds) such that those roots lie within the unit circle.
For $\lambda\in\mathbb{R}$ these conditions can be obtained directly from Jury criterion.
For $\lambda\in\mathbb{C}$, I can prove that this also holds as $\epsilon\rightarrow 0$ but I cannot find an exact bound. My approach is to prove that for the root close to 1 it holds true that $$z=1-\epsilon\lambda+O(\epsilon^2)$$ and therefore $$|z|^2=1-2Re(\lambda)\epsilon+O(\epsilon^2)$$ which results in $|z|<1$.
Completing the square gives:
$$ \left(z-\frac{1}{2}\right)^2 = \frac{1}{4} - \epsilon \lambda \;\;\implies\;\;\left|z-\frac{1}{2}\right| = \sqrt{\left|\frac{1}{4} - \epsilon \lambda\right|} $$
Therefore the roots lie on the circle of radius $\displaystyle\,r=\sqrt{\left|\frac{1}{4} - \epsilon \lambda\right|}\,$ centered at $\,\dfrac{1}{2}\,$, and a sufficient condition for them to be inside the unit circle is that $\,r \lt \dfrac{1}{2}\,$:
$$ r^4 \lt \frac{1}{16} \;\;\iff\;\; \left(\frac{1}{4} - \epsilon \lambda\right)\left(\frac{1}{4} - \epsilon \bar \lambda\right) \lt \frac{1}{16} \;\;\iff\;\;16|\lambda|^2 \cdot \epsilon^2 - 8 \operatorname{Re}(\lambda)\cdot\epsilon \lt 0 $$
Given that $\,\operatorname{Re}(\lambda) \gt 0\,$ the latter gives the sufficient condition $\,0 \lt \epsilon \lt \dfrac{\operatorname{Re}(\lambda)}{2 |\lambda|^2}\,$.