Let $z=x+yi$ and $a\ge1$ for $a,x,y \in \mathbb{R}$. We have the inequality pair: $|z+a| \le a \Rightarrow |(x+a)+yi| = \sqrt{(x+a)^2+y^2} \le a \Rightarrow (x+a)^2+y^2 \le a^2$ (i); $|z^2+a| \le a \Rightarrow |(x^2-y^2+a) + 2xyi| = \sqrt{(x^2-y^2+a)^2 +(2xy)^2} \le a \Rightarrow (x^2+y^2)^2+2a(x^2-y^2) \le 0$ (ii) and solving (ii) for $a$ gives us: $a \ge \frac{(x^2+y^2)^2}{2(y^2-x^2)} =1 \Rightarrow (x^2+y^2)^2 - 2y^2+2x^2=0 \Rightarrow y^4 - (2-2x^2)y^2 +(x^4+2x^2) = 0$; or $y^2 = \frac{(2-2x^2) \pm \sqrt{(2-2x^2)^2-4(1)(x^4+2x^2)}}{2} = (1-x^2) \pm \sqrt{1-4x^2}$ (iii). We require the discriminant in (iii) to be non-negative, or $-\frac{1}{2} \le x \le \frac{1}{2}$ (iv). SInce (i) is a circle centered at $(x,y) = (-a,0)$ and radius $a$, then $-2a \le x \le 0$ (v) $\Rightarrow x \in \left[-\frac{1}{2},0\right]$ satisfies both (iv) and (v). We also require (iii) to be entirely non-negative as well, or: $(1-x^2) + \sqrt{1-4x^2} \ge 0 \Rightarrow 0 \ge x^4+2x^2 \Rightarrow x=0$ (vi); $(1-x^2) - \sqrt{1-4x^2} \ge 0 \Rightarrow 0 \le x^4+2x^2 \Rightarrow x \in \mathbb{R}$ (vii). By (vi) we obtain the points $(x,y) = (0,0); (0, \pm \sqrt{2})$, of which only $(0,0) \in$ (i). However, we require $x \neq \pm y$ for the condition $a \ge \frac{(x^2+y^2)^2}{2(y^2-x^2)} =1$, which eliminates all of these specific points. This ultimately leaves us the set of points: $\Omega = \left\{(x,y): x \in \left[-\frac{1}{2}, 0\right); y \in [-\sqrt{a^2-(x+a)^2}, \sqrt{a^2-(x+a)^2}] \right\}$; of which $\Omega \subset$ (i). Taking the extreme point $x = -\frac{1}{2}, y = \pm\sqrt{a^2-(a-1/2)^2}$, we obtain $|z| = |x+yi| = \sqrt{x^2+y^2} = \sqrt{\frac{1}{4}+ a^2 - \left(a^2-a+\frac{1}{4}\right)} = \sqrt{a} \le a \Rightarrow a \in [1,\infty)$. QED