Why-arg-z-arg-z-2kpi-k-Z-Shouldn-t-it-be-always-0-

Question Number 19250 by Tinkutara last updated on 07/Aug/17

Why \arg (z) + \arg (\bar{z}) = 2 k π, k \in Z ?

{Shouldn}^{'} t it be always 0 ?

Commented by ajfour last updated on 07/Aug/17

yes, so do i think .

but if z = - 1 ?!

Commented by Tinkutara last updated on 08/Aug/17

Thanks Sir! It would be 2 π .