Is $|z^2 + 1| > |z|^2 - 1$?

Question

I am often confused while using complex number formula involving comparisons.

It is known that $|z^2 - a^2| > |z|^2 - a^2$. But is $|z^2 + 1| > |z|^2 - 1$?

Where $z$ is a complex number.

Also, please suggest some proofs so that it is easy to remember such formulae.

Answer 1

Recall $|a+b | \leq |a| + |b| $, the famous triangle inequality. WE have

$$ |z|^2 = |z^2+1-1| \leq |z^2 +1 | + |-1| = |z^2 + 1 | + 1 $$

Answer 2

By the triangle inequality $$|z^2+1|+1=|z^2+1|+|-1|\geq|z^2+1-1|=|z|^2.$$ The equality occurs for $z=i$.

By the way, $$|z^2+1|>|z|^2-1$$ is wrong for $z=i$.

Answer 3

In general, $$ |a - b| \geq ||a|-|b||$$ for any $a,b \in \mathbb{C}$. (Or any $a, b\in V$ a normed vector space.) This is the reverse triangle inequality. Intuitively, if you have vectors of fixed length, the way to make their difference as small as possible is to point them in the same direction.

Substituting $-b$ for $b$ also gives $|a + b| \geq ||a| - |b|| \geq |a| - |b|$. Now make the appropriate substitutions for your problem and you are done.