Let a, b be real numbers such that the complex roots ${\mathrm{z}}_1,{\mathrm{z}}_2,{\mathrm{z}}_3$ of the equation ${\mathrm{x}}^3+{\mathrm{ax}}^2+\mathrm{bx}-1=0$ satisfy $\left|{\mathrm{z}}_1\right|\ge 1,\left|{\mathrm{z}}_2\right|\ge 1,\left|{\mathrm{z}}_3\right|\ge 1$ Then the value of a+b=

By Vieta's relations, we have ${\mathrm{z}}_1{\mathrm{z}}_2{\mathrm{z}}_3=1$.
But then $\left|{\mathrm{z}}_1{\mathrm{z}}_2{\mathrm{z}}_3\right|=\left|{\mathrm{z}}_1\right|·\left|{\mathrm{z}}_2\right|·\left|{\mathrm{z}}_3\right|\ge 1$ implies $\left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|=\left|{\mathrm{z}}_3\right|=1$.
Let f(x) denote the given cubic. Since f is cubic, it must have at least one real root (this is true of any polynomial of odd degree) - this root must be $\pm 1$.

Case 1. If all three roots are real, they must be 1,1,1 or -1,-1,1, in each case implying a+b=0 by Vieta's relations.

Case 2. Suppose one root, say ${\mathrm{z}}_1$, is real. By the complex conjugate root theorem the other two must satisfy ${\mathrm{z}}_2=\mathrm{\alpha },{\mathrm{z}}_3=\overline{\mathrm{\alpha }}$ for some complex number $\mathrm{\alpha }$ such that $\left|\mathrm{\alpha }\right|=1$. Then we have $={\mathrm{z}}_1{\mathrm{z}}_2{\mathrm{z}}_3={\mathrm{z}}_1\mathrm{\alpha }\overline{\mathrm{\alpha }}=z_1|\alpha {|}^2=z_1$

Now we compute with Vieta's relations. We have $-\mathrm{a}={\mathrm{z}}_1+{\mathrm{z}}_2+{\mathrm{z}}_3=1+\mathrm{\alpha }+\overline{\mathrm{\alpha }}=\mathrm{\alpha }\overline{\mathrm{\alpha }}+\mathrm{\alpha }·1+\overline{\mathrm{\alpha }}·1={\mathrm{z}}_2{\mathrm{z}}_3+{\mathrm{z}}_1{\mathrm{z}}_2+{\mathrm{z}}_3{\mathrm{z}}_1=\mathrm{b}$, implying a+b=0 as desired.