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I encountered following problem enter image description here

and I solved it by using the hint provided. Thinking of it I noticed that I am able to solve it even if I use the following function: $$ F(z)=1/f(1/z)),\quad |z|> 1$$ $$ =f(z) , \quad |z|\leq 1 $$

What is the problem if I use this function to solve the problem? I can extend it to the whole $\mathbb{C}$ as well: I know that analytic continuation of any function is unique, but I am thinking where is problem if I choose to use this function.

Any Help will be appreciated.

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To be continuous on unit circle. when $|z|=1$ then $\bar{z}=\dfrac{1}{z}$ therefore on unit circle $$\dfrac{1}{\overline{f(1/\bar{z})}}=\dfrac{1}{\overline{f(z)}}=f(z)$$ with definition $f(z)$ in $|z|<1$ and $\dfrac{1}{\overline{f(1/\bar{z})}}$ in $|z|>1$. in your case this continuation will not be continouess.

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  • $\begingroup$ Ohh THanks a lot Sir. Is it possible to show Function that was given by me is not continuous ? $\endgroup$ – MathLover Sep 20 '18 at 13:21
  • $\begingroup$ Your assertion valid if $f(1/z)=1/f(z)$ where $z=e^{it}$. is it hold? in your case we have to $f(e^{-it})f(e^{it})=1$ $\endgroup$ – Nosrati Sep 20 '18 at 13:27

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