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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?

2 Answers 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.2012-12-13
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    @Arthur I saw. : )${}$2012-12-13
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    But this is just pushing the calculations to complex polynomials and $z \mapsto |z|$ being continuous.2012-12-13
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    @ronno Yes, if I understand you correctly that's what my answer is saying.2012-12-13
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    As in, you'd have to do similar computations to show these facts.2012-12-13
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    @ronno You get for free that $f(x) =x$ is continuous. Then you show that if $f,g$ are so is $f \cdot g$. I think I would not call it similar although I can see why one might.2012-12-13
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    I meant, we can always specialize the computations we use to show the continuity in the general case to this specific function. General computations often do turn out to be easier, if we can extract out the right features. Bad phrasing on my part, I guess.2012-12-13
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    @Matt N. Ah, very nice! Thank you!2012-12-13
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    @user52991 Glad to help. : )2012-12-13
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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 $\delta + \delta(\delta + |z|) <1 - |z+z^2|$.