What kind of inequality is this?

Question

What kind of inequality is this?

$|z^2 + 1|$ is greater than or equal to $|z|^2 - 1$

why $-1$? , and why not $-100\,000$?

does $z$ have to be a complex variable?

correction : sorry it was not $|z^2|$, it was $|z|^2$

Answer 1

If $z,w\in\Bbb C$, then $|z-w|\geqslant\bigl||z|-|w|\bigr|\geqslant|z|-|w|$. So, in particular,$$|z^2+1|=|z^2-(-1)|\geqslant|z^2|-1=|z|^2-1.$$

Answer 2

In the reals,

$$z^2+1\ge z^2$$ holds (and the modulus is not even necessary).

But in the complex, a counter-example is

$$|i^2+1|=|i^2|-1=0.$$

---

Let $z^2=a+ib$ and let us look for the minimum of

$$\sqrt{(a+1)^2+b^2}-\sqrt{a^2+b^2}.$$

We cancel the gradient, $$\begin{cases}\dfrac{a+1}{\sqrt{(a+1)^2+b^2}}=\dfrac{a}{\sqrt{a^2+b^2}}, \\\dfrac{b}{\sqrt{(a+1)^2+b^2}}=\dfrac{b}{\sqrt{a^2+b^2}}\end{cases}$$

and this implies $b=0$, and $z^2$ is real. The function reduces to

$$|a+1|-|a|$$ which has the minimum $-1$. Hence

$$|z^2+1|\ge|z^2|-1$$ is tight.