Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that the subset $U = \{z : |z+ z^2|<1\}$ is open in the $\mathbb{C}$. This seems to be a simple question. But I am not getting anywhere with it.

What I have tried so far is this. If $w$ is in $U$, then I need to find $r >0$ such that the $B$ = ball centred at $w$ with radius $r$, lies inside $U$.

Now, let $w_1 \epsilon\ B$, then $|w_1 + w_1^2| \leq|w_1 - w| + |w + w^2| + |w^2 - w_1^2|$. So if I let $|w+w^2| = \delta <1$, then $|w_1 + w_1^2| \leq r + \delta + |w^2 - w_1^2|$.

I could possibly choose $r < 1-\delta$. But that does not help me do away with the last term on the right hand side of the inequality right? Or am I missing something obvious?

share|cite|improve this question

It is easier if you use that if $f$ is continuous then its inverse maps open sets to open sets. The polynomial $p(z) = z + z^2$ is continuous. The absolute value function $|\cdot|$ is also continuous. Hence $f(z) = |p(z)|$ is a continuous function $f: \mathbb C \to \mathbb R$.

And $U = f^{-1}((-1,1))$ is the inverse image of an open set and hence open.

share|cite|improve this answer
I was just about to post the same answer myself (as a comment, though). You had me beat by seconds. – Arthur Dec 13 '12 at 18:10
@Arthur I saw. : )${}$ – Rudy the Reindeer Dec 13 '12 at 18:10
But this is just pushing the calculations to complex polynomials and $z \mapsto |z|$ being continuous. – ronno Dec 13 '12 at 18:12
@ronno Yes, if I understand you correctly that's what my answer is saying. – Rudy the Reindeer Dec 13 '12 at 18:13
@Matt N. Ah, very nice! Thank you! – user52991 Dec 13 '12 at 18:23

The argument from continuity is much better. But if you're arguing directly from the definition, suppose that $|z + z^2| < 1$. We need to show that if $|w - z| < \delta$, then $|w + w^2| < 1$. But $|w + w^2| < |w - z| + |z + z^2| + |w^2 - z^2|$.

We have good control over $|w-z|$, and $|z+z^2| < 1$ already. So we just need to fiddle a little room into $|w^2 -z^2| = |(w+z)(w-z)| < \delta |w+z|$.

Now $|w| - |z| \le | |w| - |z|| \le|w-z| < \delta$, so $|w| < \delta + |z|$.

This gives us that $|w^2 - z^2| < \delta(\delta + 2|z|)$.

So just pick $\delta$ so that $\delta + \delta(\delta + |z|) <1 - |z+z^2|$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.