Can we deduce that $|z|≥|a|-|b|$ Let us consider three complex numbers $z,a,b$ such that the equality $z=a-b$ holds true. Can we deduce that $$|z|≥|a|-|b|$$

Yes.

$$ a = b + z $$ so $$ |a| = |b + z| \leq |b| + |z|. $$

In fact, even more is true! Since $b = a - z$, you can repeat this, $$ |b| = |a - z| \leq |a| + |z| $$

and combining these two, you have

$$ |z| \geq ||a| - |b|| $$