I know that a continuous periodic function must be bounded because if a function is continuous and periodic, its graph will have to turn at certain points to reattain the values and hence, it cannot be unbounded. I tried using the concept of reimann sheets but that was not the case. Please help mei out.
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Put simply, they are periodic in one direction and unbounded in the other direction. Specifically, their values repeat if you shift by $2\pi$ along the real axis, but increase unboundedly along the complex axis.
Notice that the argument you outline in your first sentence only works in $\mathbb R^1$.
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$\begingroup$ Okay. So can we put it in this way- 2π is the period along real axis and there's no period along the imaginary axis. And hence, they are unbounded? $\endgroup$ Commented Jun 4 at 17:11
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$\begingroup$ Yes, mostly. It's possible for a function to not be periodic but still stay bounded, so that "hence" in your last sentence isn't technically valid. But for these particular functions, they are unbounded along the imaginary axis, (and thus cannot be periodic along that axis). $\endgroup$ Commented Jun 4 at 17:19
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1$\begingroup$ Also, if a complex function has two "independent" periods, things can get very interesting, and if you know a bit of complex analysis you could go look up "doubly periodic functions" to learn more. $\endgroup$ Commented Jun 4 at 17:22
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