Describe the closure of -pi 0 arg(z) subset of all real numbers -- arg(z) = all real numbers but I don't know if that describes the closure since last time when describing the closure it would be Cl(E)={zεC :-pi

Question

Describe the closure of -pi 0 arg(z) subset of all real numbers --> arg(z) = all real numbers but I don't know if that describes the closure since last time when describing the closure it would be Cl(E)={zεC :-pi

anonymous · Answer

@UnkleRhaukus @eliassaab @eseidl @experimentX @phi

experimentx · Answer

I think it might be entire R. even not considering it's periodicity, consider z = (some number)^(1/n) with n-> infinity ... it has a branch point at z=0. the argument range from 0 to infinity. but I am not so sure.

anonymous · Answer

That helps! I was thinking it was every real number excluding when z falls on the x axis since -pi and pi are not included in the set, so something like...

Cl(E)={zεC : arg(z)=Cε∏+2k∏, k ε Z} ?