Let $f$ be an analytic function defined on the unit disc in $\mathbb C$ .Then $f$ is constant if

$1.f \left(\dfrac{1}{n} \right)=0$

$2.f(z)=0$ for all $|z|=\dfrac{1}{2}$

$3.f \left(\dfrac{1}{n^2} \right)=0$

$4.f(z)=0$ for all $z\in (-1,1)$

My try:

Since in 1 and 3 ;$\dfrac{1}{n} $ and $\dfrac{1}{n^2} $ has limit point 0 which lies in the unit disc so $f(z)\equiv 0$ and hence constant and in $4$ $f$ is zero on a connected open set and hence zero identically .

So 1,3,4 are the correct options


(2) is also right, because $\{z:|z|=1/2\}$ has a limit point inside the unit disc.

  • $\begingroup$ Consider the set A= {z:|z|=1/2}. Isn't every point of A a limit point of A, which is a subset of the unit disc? $\endgroup$ – Somabha Mukherjee Jun 22 '15 at 4:49
  • $\begingroup$ @learnmore Pick any point on your circle. Can you see that this point is the limit of distinct circle points? $\endgroup$ – Nicolas Bourbaki Jun 22 '15 at 4:49
  • $\begingroup$ @learnmore: Just pick points that are on the circle but closer and closer to the desired limit point. $\endgroup$ – user21820 Jun 22 '15 at 5:03

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