let f be a meromorphic function on $\Bbb C$ such that $\lvert f(z)\rvert \ge \lvert z\rvert$ at each point z where f is holomorphic.Then which of the following are true

(1)the hypothesis is contrary and so no such function exists.

(2)there is a unique function satisfying the above conditions.

(3)such an f is an entire function

(4)there is an A$\in \Bbb C$ such that $\lvert A\rvert \ge$ 1 and f(z)=Az $\forall z\in \Bbb C$

i tried the following:

from given ,

$\lvert f(z)\rvert \ge \lvert z\rvert$ $\forall z$ where f(z) is holomorphic

so,$\lvert \frac{f(z)}{z}\rvert \ge 1$

now, i can apply little picard theorem after this if $\frac{f(z)}{z}$ is entire for which there is no guarantee....

so how to proceed after this step...

any help would be appreciated...

correct choices are(3) and (4)....

  • $\begingroup$ @Ravi:thanks for the examples..but,what i dont understand is:how can a meromorphic function under the assumptions of the above problem needs to be an entire function $\endgroup$ – Abhishek Shrivastava Oct 25 '16 at 6:50

You probably want something about the domain of $f$ being connected otherwise there are tons of such possible functions.

Clearly it is neither $1$ nor $2$ since both $f_1(z)=z$ and $f_2(z)=2z$ are examples of such functions. If $4$ were true, it would imply $3$ so you only need to prove $4$.

To prove $4$, consider the function $\frac{z}{f(z)}$ and apply the Schwarz lemma.

  • $\begingroup$ :how can we apply schwarz theorem without domain being open unit disk??please elaborate $\endgroup$ – Abhishek Shrivastava Oct 25 '16 at 17:14
  • $\begingroup$ First do it on the unit disk and then use the unique extension property of meromorphic function: $\endgroup$ – Ravi Oct 25 '16 at 19:03
  • $\begingroup$ would be really thankful if you could answer in detail $\endgroup$ – Abhishek Shrivastava Oct 25 '16 at 19:04

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