If $|z^2 + 2019| < 2019$ prove that $|z + \sqrt{2019}| > 31$ I got this from today's test.
Let $z\in \mathbb{C}$.
If $|z^2 + 2019| < 2019$ prove that  $|z + \sqrt{2019}| > 31$
I tried triangle inequality, but doesn't work. I also tried using contradiction, that is $|z + \sqrt{2019}| \leq 31$ then letting $z = x + iy$ but then turn to a mess. So far I've got $$(x^2 + y^2)^2 + 4038(x^2 - y^2) < 0$$ from the assumption.
Any hint?
Thanks!
 A: The solution follows from the following lemma.
Lemma. For $z\in\mathbb C$, we have that
$$|z^2+1|<1\qquad\text{implies}\qquad |z+1|>\frac{1}{\sqrt 2}.$$
The solution follows easily once we have the lemma, since we may introduce a positive real $r>0$ and replace $z$ with $z/r$ to deduce that
$$
|z^2+r^2|<r^2\qquad\text{implies}\qquad |z+r|>\frac{r}{\sqrt 2}.
$$
Taking $r=\sqrt{2019}$ and noting that $\sqrt{2019/2}>31$ yields the solution.
Now for the proof of the lemma, observe that
$$
|z^2+1|^2=(|z|^2-1)^2+(z+\bar{z})^2\geq \frac{(|z|^2+z+\bar{z}-1)^2}{2}=\frac{(|z+1|^2-2)^2}{2},
$$
and thus $|z^2+1|<1$ implies that $|z+1|^2>2-\sqrt{2}$, which is a slightly stronger bound than in the lemma (since $2-\sqrt{2}>1/2$).
Side note. Putting the tight bound through, we see that the best constant in place of $31$ is 
$$
\sqrt{2019}\cdot\sqrt{2-\sqrt 2}\approx 34.39
$$
To see that this truly the best constant, it is necessary to consider the equality case in the inequality we used during the proof of the lemma. In fact, we can tighten the inequality to an identity
$$
2|z^2+1|^2=\bigl(|z+1|^2-2\bigr)^2+\bigl(|z-1|^2-2\bigr)^2,
$$
which illuminates what is going on.
A: Here is a geometrical proof.
Let us prove the (equivalent) contraposal assertion :
$$|z + \sqrt{2019}| \leq 31 \ \  \implies \ \ |z^2 + 2019| \geq 2019\tag{1}$$
Otherwise said, in terms of disks $D((a;b),R)$ where $(a,b)$ is the center and $R$ the radius, let us prove that : 
$$z \in \underbrace{D((-\sqrt{2019};0),31)}_{D_1} \ \  \implies \ \ z^2 \notin \underbrace{D((-2019;0),2019)}_{D_2}\tag{2}$$
Consider the figure below featuring disk $D_1$ only, where $OB$ and $OC$ are tangent to $D_1$.
Let $$\alpha:=\arcsin(\dfrac{CB}{OC})=\arcsin(\dfrac{31}{\sqrt{2019}})=0.7614<\dfrac{\pi}{4}.$$
Let us take $z \in D_1$. As $D_1$ is entirely included into the angular sector defined by $AOB$, we have :
$$\pi-\alpha\leq \arg(z)\leq\pi+\alpha.$$
As a consequence, 
$$2(\pi-\alpha)\leq \arg(z^2)\leq 2(\pi+\alpha).$$
which amounts to say, mod $2 \pi$  :
$$-2\alpha\leq \arg(z^2)\leq 2\alpha,$$
Therefore, due to the fact that $0< 2 \alpha <\pi/2$, $z^2$ belongs to the half plane $H$ defined by $Re(z)\geq 0$, thus cannot belong to $D_2$ which is included in the complementary set of $H$, reaching the desired RHS of (2).
Remarks : 
1) $\arcsin(32/\sqrt{2019})= 0.7926 > \dfrac{\pi}{4}$ showing that $31$ has been chosen in a "tight" way.
2) The very particular case $z=0$ which belongs to $H$ and $D_2$ cannot hold because $0$ doesn(t belong to $D_1$.

