Show that there is no function $f$ that is analytic in punctured unit disc and $f'$ has a simple pole at $0$.

Question

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.

Yes, that is correct. But you should also say why it is correct. — Daniel Fischer, Apr 18 '17 at 15:55
@DanielFischer since $a_{-1}$ is not zero and $\int \frac{1}{z} dz=2\pi i$ — Eklavya, Apr 18 '17 at 16:00
Yes, it is. Note that what you have shown is essentially that "the residue of a derivative at an isolated singularity is $0$". That's a useful thing sometimes. — Daniel Fischer, Apr 18 '17 at 16:05

score 4 · Accepted Answer · answered Apr 18 '17 at 17:51

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!

@error404 Thank you. Much appreciated. – Mark Viola Jun 30 '18 at 13:10 — Mark Viola, Jun 30 '18 at 13:10

Show that there is no function $f$ that is analytic in punctured unit disc and $f'$ has a simple pole at $0$.

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