I encountered following problem 
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.
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.