Let $f$ be holomorphic in $\mathbb{C}\setminus\{i,2i\}$. Let $f$ be holomorphic in $\mathbb{C}\setminus\{i,2i\}$. Show that if $f$ has an non avoidable singularity in $z = i$ and $z = 2i$, then, the Laurent series of $f$ in $\{1 <|z| < 2 \}$ has infinite positive and negative terms.
I've know that if $f$ has a singularity in $z_0$, then:


*

*If is avoidable, then the Laurent series of $f$ in $z_0$ has all positive terms

*If is a pole, then the Laurent series of $f$ in $z_0$ has finite negative terms

*If is essential, then the Laurent series of $f$ in $z_0$ has infinite negative terms.


In this case, $z_0 = 1$, but I don't have to evaluate in $z_0$, but in $z_1 = i$ and $z_2 = 2i$. I tried by the absurd. If 
$$\sum_{n=k_0}^\infty a_n z^n$$ is the Laurent series in the domain, then I should show that 
$$\sum_{n=k_0}^\infty a_nw^n$$
converges to a number for $w=i$ and $w=2i$. I'm assuming that the case que has finite positive terms is analogous
 A: You can write $f(z)= \int_C \frac { f(w)}{w-z} dw$   for region $1<|w|<2$
Note, that in this region $f(w)$ is holomorphic. 
$f(z)= \int_C \frac { f(w)}{w-z} dw =  \int_{C_1}{ \frac { f(w)}{w-z} dw} - \int_{C_2} \frac { f(w)}{w-z} dw$  where $C_1$ is boundary of $|w|<2$ &  $C_2$ is boundary of $|w|>1$ 
$f(z)= \int_{C_1}{ \frac { f(w)}{w(1-z/w)} dw} + \int_{C_2} \frac { f(w)}{z(1-w/z)} dw$ 
$= \int_{C_1}{ \frac { f(w)}{w}  [\sum_{k=0}^{\infty} {(z/w)^k}] dw} + \int_{C_2} \frac { f(w)}{z}  [\sum_{k=0}^{\infty} {(w/z)^k}] dw $
$= \sum_{k=0}^{\infty} {\int_{C_1}{ \frac { f(w)}{w}  {(z/w)^k} dw}} + \sum_{k=0}^{\infty} {\int_{C_2} \frac { f(w)}{z} {(w/z)^k} dw }$ 
$ \int_{C_2} w^kf(w) dw$ can't be zero for all but finitely many k, otherwise by Morera's theorem, $f(w)$ will be holomorphic in the region $|w| \le 1 $
Edit : To make it more clear, you can put powers of $z$ out of integration.
It can also be manipulated in this way : 
$f(z)= \sum_{k=1}^{\infty} a_{-k}\frac{1}{(z-i)^k} + \sum_{k=1}^{\infty} a_k(z-i)^k$   for  $|z|<2$
Now, for $|z|>1$, you clearly have  $\sum_{k=0}^{\infty} (\frac iz)^k$ convergent. 
So, $f(z)= \sum_{k=1}^{\infty} a_{-k}\frac{1}{z(1-i/z)^k} + \sum_{k=1}^{\infty} a_k(z-i)^k= \sum_{k=1}^{\infty} [\frac{a_{-k}}{z} (\sum_{j=0}^{\infty}(i/z)^j )] + \sum_{k=1}^{\infty} a_k(z-i)^k $
which will clearly give infinitely many negative powers of $z$
A: Hint: Suppose the Laurent series of $f$ has only finitely many negative power terms. Then in the given annulus we would have
$$f(z)=p(1/z)+g(z),$$
where $p$ is a polynomial and $g$ is holomorphic in $D(0,2).$ That implies $f$ is behaving too well at $i$ ...
A: Let $$f(z) = \sum_{n=k_0}^\infty a_nz^n = \sum_{n=k_0}^{-1} a_nz^n + \sum_{n=0}^\infty a_nz^n = A(z) + B(z)$$ be the Laurent series of f in $1 < |z| < 2$, where A and B are the sums. 
Then, A is a finite sum that only has problems in 0. Then A(i) is a number. 
B is holomorphic in its domain for beign a sum with no negative terms. Then, if I take an $r \in \mathbb{R}$ such that 1 < r < 2, for Cauchy-Hadamard we have that B converges for every $z$ that $|z| < r$. In particular, $|i|=1<r$. The B(i) is a number.
The $f(i)=A(i)+B(i)$ is a number so the singularity was avoidable, which is an absurd.
In the case we have infinite negative terms, but finite positive, it follows something similar, with the diference that the sum between $-\infty$ and $-1$ will converge when $|z|>1$, so, in that case, $f(2i)$ will be an avoidable singularity
