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I am trying to solve the following problem:

Prove that the following inequality defines an open subset of $\mathbb{C}$: $\lvert z + z^2 \rvert < 1$.

I tried taking a ball around each point $z$ and proving that every point in the ball satisfies the inequality. I also tried to prove that the complement is closed, but no luck. Please if you can help me that would be awesome. Thanks!

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I thought I'd provide a pedestrian solution in case you have not done (or choose not to use) the topological argument from the other solutions.

Note:

$$|z+h+(z+h)^2|=|z+z^2+h+2zh+h^2|\le|z+z^2|+|h||1+2z+h|\lt|z+z^2|+|h|(2+2|z|)$$

whenever $|h|\lt 1$. Therefore, if $z$ is in your subset, let's set $\epsilon=1-|z+z^2|$, and then choose $h\in\mathbb C$ such that $|h|\lt\frac{\epsilon}{2+2|z|}, |h|\lt 1$, and you will have:

$$|z+h+(z+h)^2|\lt|z+z^2|+\epsilon=1$$

i.e. the whole circle around $z$ and with radius $\min(\frac{\epsilon}{2+2|z|},1)$ is contained in your subset.

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$f(z)=|z+z^2|$ is continuous, so $f^{-1}(-\infty,1)$ is open.

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Note that the function $$f(z) = |z+z^2|$$ is continuous. What can you tell about the continuous pre-image of an open set?

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