TITLE: Show that there is no function $f$ that is analytic in punctured unit disc and $f'$ has a simple pole at $0$.
QUESTION [5 upvotes]: Show that there is no function $f$ that is analytic in punctured unit disc and $f'$ has a simple pole at $0$.
Let such function exist.And I have $\int( f'(z)-a_{-1}\frac{1}{z})dz=0 $ over some closed circle in open unit disc. Then how to proceed further?
Is this correct to say since $f(z)$ is primitive of $f'(z)$. then $\int f'(z) dz =0$ along that circle. Then it follow $\int (a_{-1}\frac{1}{z})dz=0 $ which is not true.

REPLY [3 votes]: As Daniel Fisher attested, the method in the OP is solid.  But, I thought it might be instructive to present another way forward.  To that end, we proceed.

If $f$ is analytic in the punctured unit disk, then it can be represented by its Laurent series
$$f(z)=\sum_{n=-\infty}^\infty a_nz^n \tag 1$$
Differentiating $(1)$, we find that 
$$\begin{align}
f'(z)&=\sum_{n=-\infty}^\infty n a_nz^{n-1}\\\\
&=\sum_{n=-\infty}^\infty (n+1)a_{n+1}z^{n}\\\\
&=\sum_{n=1}a'_nz^n
\end{align}$$
where $a'_n=(n+1)a_{n+1}$.  
Observing that $a'_{-1}=0$, we find that $f'(z)$ does not have a simple pole at $z=0$ and hence the residue of $f'(z)$ at $z=0$ is $0$.
We conclude that there is no function that is analytic in the unit puncture disk for which $f'$ has a simple pole at $z=0$.  And we are done!