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

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.

• Do you know the triangle inequality? This is an immediate consequence. (In both cases, you need to replace $>$ with $\geq$, by the way.) – Clayton Mar 9 at 19:10
• Strict inequality is false. $z=i$ results in "$0\gt0$". – alex.jordan Mar 9 at 19:11

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

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

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 $$z=i$$.

• Change $z=i$ to $z=\pm i$? – Barry Cipra Mar 9 at 19:35
• @Barry, $z=i$ it's enough. – Michael Rozenberg Mar 9 at 19:44
• Agreed, a single counterexample suffices. But it seems simple enough here to include both. – Barry Cipra Mar 9 at 19:53

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.