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2
votes
4answers
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$\{z\in C:|z| = |\operatorname{re}(z)| +|\operatorname{im}(z)|\}$ open or closed [duplicate]
Possible Duplicate:
Proving that a complex set in open/closed/neither and bounded/not bounded
I think $\{z\in C:|z| = |\operatorname{re}(z)| +|\operatorname{im}(z)|\}$ is closed. But I have ...
-4
votes
1answer
81 views
Boundedness on strips in the complex plane for functional equations [closed]
We know that the recurrence for $b>0$
(1) $f(0)=1$
(2) $f(z+1)=b{f(z)}$
has $f(z)=b^z$ as the only entire solution that is bounded on the strip $S=\{z: 0<\Re(z)\le 1\}$.
The image of $S$ ...