Consider the function $f(z)=(4z^2+1)\tanh (\pi z)$. Determine its singularities, and, in particular, the residues in the poles.
This first part should be easy: the singularities of $\tanh (\pi z)$ are in the points $\{\frac i 2 +ik:k\in \mathbb Z \}$, and they are all simple poles since the hyperbolic tangent has period equal to $\pi i$ and$$\lim _{z\to \frac i 2}\frac {e^{2\pi z}-1} {e^{2\pi z}+1}=\frac {-2}{2\pi z}=-\frac 1 {\pi z}\ .$$ However $4z^2+1$ has simple zeros in $\frac i 2$ and $-\frac i 2$, so in the end the singularities of $f$ are in $\{\frac i 2 +ik:k\in \mathbb Z, k \ne 0,-1 \}$; they are simple poles with residues $\frac {4k^2+4k} \pi$, for the corresponding $k$, because $4(\frac i 2 + ik)^2+1=-(4k^2+4k)$.
Then the exercise asks to prove that, if a function $g$ is holomorphic at infinity, then $g(z)$ is the sum of a series of the form $\sum_{j=0}^\infty a_jz^{-j} $, when $|z|$ is large enough.
I reasoned in this way: if $g(z)$ is holomorphic at infinity, it means that $G(z)=g(\frac 1 z)$ is holomorphic in $0$. So in any $z_0$ near $0$ we can write $G(z)=G(z_0)+G'(z_0)(z-z_0)+\dots$ , or equivalently $g(\frac 1 {z})=g(\frac 1 {z_0})+g'(\frac 1 {z_0})\frac 1 t+\dots$ , where $t=\frac 1 {z-z_0}$. As $z\to z_0$, $t$ should approach $\frac 1 {z_0}$, however I don't feel like this proof is correct. Can you help me with this?
Finally, the exercise asks to (1) calculate $\int_{D_R(0)} f(z)dz$, where $D_R(0)$ is the disk centered in $0$ with radius $R$, when $R$ is large enough; and (2), to study how $\frac {f(z)} {z^3}$ behaves at infinity.
Now, I thought that to calculate the integral, I could just calculate (using residues theorem) $$2 i \sum_{k=1}^\infty(4k^2+4k)+ 2i \sum _{k=2}^\infty (4k^2-4k)=16i+2i \sum _{k=2}^\infty 8k^2\ .$$ However this doesn't seem to have much sense to me, because as $R$ goes to infinty it doesn't converge; so, it's very likely that I'm wrong somewhere, even if I don't see where. For what regards point (2), it should be immediate that the point at infinity is an essential singularity for $\frac {f(z)} {z^3}$, because otherwise it would be meromorphic on the whole Riemann sphere; however $\frac {f(z)} {z^3}$ is not rational, and this is a contradiction. Thank you in advance for any help
I think your argument for the singularities of $f(z)$ is mostly correct (as in, I think what you're trying to say is correct), however precisely what you've written doesn't quite make sense. Note that your limit $$\lim_{z \to \frac{i}{2}} \frac{e^{2 \pi z}-1}{e^{2 \pi z}+1}$$ does not exist, and the RHS of your equality contains $z$, despite taking $\lim_{z \to \frac{1}{2}}$. However, you're computing virtually the correct thing, that is, $$\tanh (\pi z) = \frac{1}{\pi} \left(z - \frac{i}{2}\right)^{-1} + \dots$$ near $z = \frac{i}{2}$, which has its pole cancelled by the corresponding simple zero of $4z^2+1$.
Personally, my preferred argument here would be that $\cosh \pi z$ has simple roots, so $\tanh \pi z$ has simple poles at $z_k = i \left( \frac{1}{2} + k\right)$ for $k \in \mathbb{Z}$, with those at $z_0$ and $z_{-1}$ cancelled by the simple zeros of $4z^2+1$. For the remaining $z_k$, \begin{align*} \operatorname{res}\left(f(z), z_k \right) &= \lim_{z \to z_k} (z-z_k) f(z) \\ &= (4z_k^2+1) \sinh (\pi z_k) \cdot\lim_{z \to z_k} \frac{z-z_k}{\cosh \pi z_k} \\ &= (4z_k^2+1) \sinh (\pi z_k) \cdot \lim_{z \to z_k} \frac{1}{\pi \sinh \pi z_k} \\ &= \frac{4z_k^2+1}{\pi} = \frac{4k(k+1)}{\pi} \end{align*} where we applied L'Hôpital to deduce the value of the limit (which also shows we do indeed have removable singularities at $z_0$ and $z_{-1}$).
I'm going to state that holomorphic means analytic and the result essentially falls out immediately. It's entirely possible that this distinction hasn't even been noted to you, and often these terms get used interchangeably.
If $g$ is holomorphic at infinity, then it is holomorphic at $u \equiv \frac{1}{z} = 0$, and so analytic in some disc $\lvert u \rvert < \frac{1}{R}$. Then, within this disc, we can expand it as a Taylor series about $u= 0$, $$g|_u = \sum_{j=0}^\infty a_j u^j$$ Translating this back to $z$, we have that for $\lvert z \rvert > R$, $$g(z) = \sum_{j=0}^\infty a_j z^{-j}$$
For $R$ sufficiently large (note not in the limit $R \to \infty$, however), say with $N < R < N + 1$ for some $N \in \mathbb{N}$, we use the residue theorem. Recall there are poles for $k \in \mathbb{Z} \setminus \lbrace 0, -1 \rbrace$, so we have $$I = \int_{D_R(0)} f(z) \; dz = 2 \pi i \sum_{k = -N}^N \frac{4k(k+1)}{\pi}$$ where we've included the sum over $k = 0, -1$ since these terms do not contribute. Noting that $$\sum_{k=-N}^N k = 0, \quad \sum_{k=1}^N k^2 = \frac{N(N+1)(2N+1)}{6}$$ we can simplify $I$ to $$I = 8i \sum_{k=-N}^N k^2 = 16 i \sum_{k=1}^N k^2 = \frac{8i}{3} \cdot N(N+1)(2N+1)$$ For the behaviour of $f(z) z^{-3}$ as $z \to \infty$ - well, this is somewhat more subjective of a question. The leading order behaviour (due to the $4z^2+1)$ is going to be given by studying $$g(z) = \frac{4 \tanh(\pi z)}{z}$$ It should be clear that this is not holomorphic at $\infty$, since there are infinitely many poles - at each $z_k$ - and we have $\lvert z_k \rvert = k + \frac{1}{2} \to \infty$.
Moreover, if we consider $u = z^{-1}$, then we see that the singularities accumulate at $0$, i.e. every region $\lvert u \rvert < \delta \ll 1$ contains a countably infinite number of poles. That is, $u = 0$ is not an isolated singularity.