# Open subset of Complex plane

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?

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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 $\delta + \delta(\delta + |z|) <1 - |z+z^2|$.

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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.

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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