# Convergence Domain $\{|z|+|w| < 1\}$

Construct a power series whose domain of convergence is $\{(z, w) \in \mathbb{C}^2 : |z|+|w| < 1\}$

Having a bit of trouble with this problem, and was wondering if anyone had any ideas. There's a similar problem here: Convergence domain: $\{(z,w):|z|^2+|w|^2 < 1\}$, but this domain is a bit trickier.

Any help would be appreciated. Thanks

$\sum_{k=0}^\infty (z + e^{ik} w)^k$ from the answer in the question you linked (just with the squares removed) should work, for the same reason.
• I can't seem to show it going the other way, that is, when we have $|z| +|w| >1$ and have to show $|z + e^{ik}w| > 1$. Not entirely sure how to do this in the earlier (squared) case either... – K.Reeves Mar 19 '16 at 11:53
• @K.Reeves The point is that in that case, while there might be $k$ such that $|z + e^{ik}w|< 1$, and there might even be a lot of them, there are infinitely many times when this is not true. This is because the numbers $e^{ik}$ are dense in the unit circle. Therefore, given any angle $\theta$ and any margin $\epsilon_\theta>0$, the direction of $we^{ik}$ from the origin in the complex plane will be within $\epsilon_\theta$ of $\theta$. Including close enough to the direction of $z$ for their sum to have length more than $1$. – Arthur Mar 19 '16 at 12:03
• Okay, I think I see what you mean. So why does making the assumption $|z|+|w|<1$ imply that $|z + e^{ik}w| < 1$ for all k? – K.Reeves Mar 19 '16 at 12:27
• @K.Reeves By the triangle inequality: $|z+e^{ik}w|\leq |z|+|e^{ik}w|=|z|+|w|$. – Arthur Mar 19 '16 at 13:28
• Right, so you can't really argue with inequality/ies that $|z+e^{ik}w| > |z|+|w| >1$? You just have to use the reasoning above? – K.Reeves Mar 19 '16 at 23:44