Roots:
$Z_1$= $\frac{v(1+ \alpha)+ \sqrt{v^2(1+\alpha)^2 -4 \alpha}}{2}$
$Z_2$= $\frac{v(1+ \alpha)- \sqrt{v^2(1+\alpha)^2 -4 \alpha}}{2}$
It is clear that $|Z_2| \leq|Z_1|$
However I'm stuck on how to begin to show $|Z_1|$ $\leq$ $1$. I was wondering if anybody can help. I would like to thank you ahead of time.
The hint that was given is:
Case I: $v^2(1+ \alpha)^2-4 \alpha \geq 0$
Case II: $v^2(1+ \alpha)^2-4 \alpha < 0$
Triangle inequality $|x+y| \leq |x|+|y|$
where $0<v<1$ and $0\leq \alpha \leq 1$
Case I: $v^2(1+\alpha)^2 - 4\alpha \geq 0$, thus $Z_1$ is a real root. Its clear that $0 \leq Z_1$, so you only need to show: $Z_1 \leq 1 \iff v(1+\alpha) + \sqrt{v^2(1+\alpha)^2 - 4\alpha} \leq 2\iff \sqrt{v^2(1+\alpha)^2-4\alpha} \leq 2 - v(1+\alpha)\iff v^2(1+\alpha)^2 - 4\alpha \leq 4- 4v(1+\alpha) + v^2(1+\alpha)^2 \iff 4v(1+\alpha) \leq 4(1+\alpha)\iff v \leq 1$ which is true by assumption.
Case II: $v^2(1+\alpha)^2 -4\alpha < 0$, then $Z_1$ is an imaginary root, thus $|Z_1| \leq 1 \iff |Z_1|^2 \leq 1 \iff |v(1+\alpha) + i\cdot \sqrt{4\alpha - v^2(1+\alpha)^2}|^2 \leq 4 \iff v^2(1+\alpha)^2 + 4\alpha - v^2(1+\alpha)^2 \leq 4 \iff \alpha \leq 1$ which is also true by assumption.
This completes the proof.