I am stuck with the statement given in the title. If $f$ is a meromorphic function, and $c\neq 0$, a complex constant, then $f(z+c)$ is also a meromorphic function of the same order. If $f$ is a periodic function of period $c$, then it is obvious. What happens when it is not periodic? Are my thoughts correct? Please help me proceed. The order of a meromorphic function is defined as $\sigma=\displaystyle{\limsup\limits_{r\rightarrow\infty}\frac{\log T(r,f)}{\log r}}$, where $T(r,f)$ is Nevanlinna the characteristic function.
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1Do you mean to ask: if $f(z)$ has a zero/pole of order $k$ (at $z=0$), then $f(z+c)$ also has a zero/pole of order $k$ (at $z=-c$)? – Keshav May 20 '20 at 04:49
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No. I am asking about the order of growth of f and f(z+c) – mathenthusiast May 20 '20 at 07:56
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1Can you please define what the order of a meropmorphic function is? – user619894 May 20 '20 at 09:53