It is given that $z_o\in \mathbb{C}$ is a point such that $|z_0-i|=2$, and $f$ is a function which is analytic in $\mathbb{C}$ \ $ \{z_0\}$. It is given that $z_o\in \mathbb{C}$ is a point such that $|z_0-i|=2$, and $f$ is a function which is analytic in $\mathbb{C}$ \ $ \{z_0\}$. It is also given that $f$ has a pole at $z_0$ and $f^{(k)}(i) \ge 0$ for each non-negative integer $k$. Prove that $z_0=2+i$
 A: Let $g(z)=f(i+z)$. Then $g$ has  a unique pole in $z=z_0-i$. We have $g^{(k)}(z)=f^{(k)}(z+i)$, hence $a_k=g^{(k)}(0)$ is real and $\geq 0$ for all $k$. Put $\displaystyle b_k=\frac{a_k}{k!}$. Then, for $|z|<2$, we have $\displaystyle g(z)=\sum b_k z^k$. Suppose now that $z_0-i\not = 2$. Then,  for $x\in [0,2[$ and $x\to 2$, $g(x)$ has a finite limit, $L=f(i+2)$ ($L $ is real positive as this is the case for $g(x)$, $x\in [0,2[$). Now clearly, as the $b_k$ are $\geq 0$, the function $g$ is increasing on $[0,2($ and bounded by $L$; we get that for all $N$, $x\in [0,2($, we have $\displaystyle \sum_{k=0}^{N}b_k x^k\leq L$, and now we let $x\to 2$; we get $\displaystyle \sum_{k=0}^{N}b_k 2^k\leq L$ for all $N$, this show that the series $b_k 2^k$ is convergent. This imply that for any $z$ with $|z|=2$, the serie $b_k z^k$ is (absolutely) convergent, and this contradict the fact that $g$ has $z_0-i$ for pole. Hence we must have $z_0=2+i$. 
If I remember well, this is only a particular case of the fact that if a power series $h$ of radius of convergence $R>0$ has positive Taylor coefficients, then the point $z=R$ is a singularity of the series. Perhaps somebody can say from which Mathematician is this theorem.  
