Just Calculate the Residues

In an exam for a course of Complex Analysis last year they asked to compute the following integral:
$$\int_{\partial B_1 (0)} \frac{dz}{(e^{2\pi z}+1)^2}.$$ To compute it, we only have to calculate the residues at $\pm \frac{i}{2}$, the zeros of $(e^{2\pi z}+1)^2$ inside $B_1(0)$.

To do this we'll calculate the relevant coefficient $d_{-1}$ of the Laurent series of $\frac{1}{(e^{2\pi z}+1)^2}$ at $\pm \frac{i}{2}$. It's easy enough. Let $f(x)=e^{2\pi z}+1$.  Let's suppose we know (And it's true that we know) how to calculate its Maclaurin series:
$$f(z)=\sum_{n=0}^\infty a_n (z-z_0)^n$$
where the $z_0$ in the right side can be replaced by $\pm \frac{i}{2}$. Let $g(z)=\frac{1}{f(z)}$ have a Laurent series expansion
$$g(z)=\sum_{n=-1}^{\infty} b_n (z-z_0)^n$$
in a neighborhood $U$ of $z_0$ (We can do this since $z_0$ is a simple zero of $f$).

Then $f(z)g(z)=1$ in $U$. With this fact we'll compute the terms $b_{-1}=\Res_{z_0} g$ and $b_0$. In the product of the series, the coefficient of $z_0$ is $1=c_{0}=a_0b_0+a_1b_{-1}$. Since $a_0=0$, then $1=a_1 b_{-1}$ so that $b_{-1}=\frac{1}{a_1}=\frac{1}{f'(z_0)}$(This fact is well known). In the same way we can see that
$$a_0 b_1+a_1 b_0+a_2 b_{-1}=0$$
which results in
$$a_1 b_0+a_2 b_{-1}=0$$
or
$$b_0=-\frac{a_2}{a_1^2}=-\frac{1}{2}\frac{f''(z_0)}{f'(z_0)^2}.$$
We have now calculated enough terms. Now, knowing that
$$g(z)=\sum_{n=-1}^\infty b_n z^n$$
we must only calculate the residues of $g(z)^2$ at $z_0$.
But, as before, the coefficient $d_{-1}$ of $z^{-1}$ will be
$$d_{-1}=b_0 b_{-1}+b_{-1}b_0=2b_0 b_{-1}=-\frac{2}{f'(z_0)}\frac{1}{2}\frac{f''(z_0)}{f'(z_0)^2}=-\frac{f''(z_0)}{f'(z_0)^3}.$$

Now we're ready. See that $f'(z)=2\pi e^{2\pi z},f''(z)=(2\pi)^2 e^{2\pi z}$, and since $e^{2\pi z_0}=-1$, $f'(z_0)=-2\pi,f''(z_0)=-4\pi^2$.

So $\Res_{z_0} \frac{1}{e^{2\pi z}+1}=d_{-1}=-\frac{-4\pi ^2}{-2\pi (4\pi ^2)}=-\frac{1}{2\pi}$.

And last, by the residue theorem,
$$\int_{\partial B_1(0)} \frac{dz}{(e^{2\pi z}+1)^2}=2\pi i \frac{-1}{2\pi}+2\pi i \frac{-1}{2\pi}=-2i.$$

About the Meromorphic functions on the Riemann Sphere

Proposition. Let $f:\hat{\bC}\to \hat{\bC}$ be a meromorphic function. Then $f$ is a rational function.

To prove this, we'll use some lemmas.

Lemma 1. Let $f$ be such a meromorphic function. Then it has only a finite number of poles.
Proof.  A meromorphic function is just a holomorphic function in $\hat{\bC}$ (minus a discrete set of poles). If $a=\infty$ then, since the set $S$ of singularities is discrete and closed, then it's discrete and compact, thus finite.

Proof of Proposition. Let $a_1,\cdots, a_n$ be the poles of $f$ and take
$$g(z)=\prod_{i=1}^n (z-a_i)^{o(a_i)} f(z).$$
Then $f(z)$ is holomorphic in $\bC$, and still meromorphic in $\hat{\bC}$. So,
$$\lim_{z\to\infty} g(z)=\in \hat{\bC}.$$
If $\lim_{z\to\infty} g(z)\in \bC$, then by Liouville theorem $g$ is constant and $f(z)=\frac{g(12)}{\prod_{i=1}^n(z-a_i)^{o(a_i)}}$.

If $\lim_{z\to\infty} g(z)=\infty$ then $\infty$ is a pole of $g$, and by the lemma shown here, $g$ is a polynomial. In both cases we're done.

About Entire Functions Which are Polynomials

Let $f$ be an entire function and suppose that $\infty$ is a pole of order $n$, so that $g(z)=f(\frac{1}{z})$ has a pole of order $n$ at $0$. Then, in a punctured neighborhood of $0$, $g$ has a Laurent series expansion
$$g(z)=\sum_{k=-n}^\infty a_n z^n.$$
Furthermore
$$\lim_{z\to\infty}g(z)=\lim_{z\to\infty} f(\frac{1}{z})=\lim_{z\to 0} f(z)$$
so
$$g(z)=\sum_{k=-n}^0 a_n z^n.$$
Therefore, in a punctured neighborhood of $\infty$, given by $|z|>R$ for some $R>0$ , $f(z)=g(\frac{1}{z})=\sum_{k=0}^n a_{-n} z^{n}$. Thus, $f$ agrees with a polynomial in a big enough set, and by the identity theorem $f$ is a polynomial. The next result follows:

Lemma. If $f$ is an entire function with $\infty$ as a pole, then $f$ is a nonconstant polynomial.

The reciprocal is true: Every nonconstant polynomial has $\infty$ as a pole.

Proposition.  Let $f,g$ be entire functions such that $f\circ g$ is a nonconstant polynomial. Then both $f$ and $g$ are polynomials.

Proof. If $f\circ g$ is a nonconstant polynomial then $f\circ g(z)\to \infty$ when $z\to \infty$. Let's suppose that $g$ is not a polynomial. Then $\infty$ is a removable singularity or an essential singularity of $g$. In the former case $g$ is entire in the extended plane, and by Liouville theorem, $g$ is constant, thus $f\circ g$ is also constant, a contradiction. So $\infty$ is an essential singularity of $g$. In this case, by Casorati-Weierstrass theorem, there exists a sequence $s_n\to\infty$ such that $g(s_n)\to a$ where $a$ is any complex number. Then $f\circ g(s_n)\to f(a)\neq \infty$, a contradiction. So, $g$ must be a nonconstant polynomial. Similarly, if $f$ is not a polynomial, and $\infty$ is an essential singularity of $f$ (If it's removable we can argue as above to show $f\circ g$ is constant) then by Casorati-Weierstrass theorem there is some sequence $s_n\to\infty$ such that $f(s_n)\to a'$. Also, by the Fundamental Theorem of Algebra, the equations $g(z)=s_n$ always have solutions, so we get some choice of sequence $r_n\to\infty$ such that $g(r_n)=s_n$, so that $f\circ g(r_n)\to a'$, another contradiction. So $f$ is a polynomial.

Note: The nonconstant condition cannot be removed: You can take any entire function as $g$ and any constant polynomial as $f$ and $f\circ g$ will be a constant polynomial.

About the singularities of $\sin \frac{1}{\cos\frac{1}{z}}$

Exercise. Find and classify the singularities of $f(z)=\sin\frac{1}{\cos\frac{1}{z}}$.

Solution. Whenever $\cos \frac{1}{z}$ is holomorphic and doesn't vanish $f(z)$ is holomorphic. Since all the zeros of $\cos z$ are its real zeros, i.e. $z=(2k+1)\frac{\pi}{2}$, then the only singularities of $f(z)$ are when  $z=0$, and when $z=\frac{1}{(2k+1)\frac{\pi}{2}}$. It may happen that one or more of them are removable, but this is not the case.

Let's first consider $z_0=0$. In this case, the singularity is not isolated.

Now consider $z_0=(2k+1)\frac{\pi}{2}$ for some $k$. Then if $z\to z_0$ along the real axis, $\cos \frac{1}{z}\to 0$ along the real axis, and if $z\to z_0$ in such a way that $\frac{1}{z}\to \frac{1}{z_0}$ along the line $\frac{1}{z_0}+ti$, then $\cos\frac{1}{z}\to 0$ along the imaginary axis. This, because $$\cos \frac{1}{z}=\cos\left(\frac{1}{z_0}+ti\right)=\cos\left(\frac{1}{z_0}\right)\cosh(t)-i\sin\left(\frac{1}{z_0}\right)\sinh(t)=\pm i\sinh(t)$$ where the sign depends on $k$.

So, in these cases, $\frac{1}{\cos\frac{1}{z}}\to \infty$ along the real axis and the imaginary axis respectively. Along the real axis, $f(z)$ oscilates, and in particular, is bounded, but along the imaginary axis, $f(z)\to\infty$. So these singularities are essential.

An Application of Schwarz Lemma

Schwarz Lemma (Theorem 4.3.4.13 of Ahlfors Complex Analysis).  Let $f$ be holomorphic on $B_1(0)$ such that $|f(z)|\leq 1$ and $f(0)=0$. Then, for all $z$, $|f(z)|\leq |z|$ and  $|f'(0)|\leq 1$. If $|f(z)|=|z|$ for some $z\neq 0$, or if $|f'(0)|=1$, then $f(z)=cz$ with a constant $c$ of absolute value $1$.
Lemma. Let $\mu(z)=\frac{z-a}{\overline{a}z-1}$ for a fixed $a\in B_1(0)$ and each $z\in \bC$. Then $\mu(B_1(0))= B_1(0)$.
Proof. First note that  $|\mu(0)|=|a|<1$.  Furthermore,
$$|\mu(1)|=\left|\frac{1-a}{\overline{1-a}}\right|=1,|\mu(-1)|=\left|\frac{1+a}{\overline{a+1}}\right|=1$$
and
$$|\mu(i)|=\left|\frac{i-a}{\overline{a}i-1}\right|=\left|\frac{i-a}{-\overline{a}-i}\right|=\left|\frac{i-a}{\overline{i-a}}\right|=1.$$
So $\mu$ is a Möbius transformation sending $\partial B_1(0)$ to $\partial B_1(0)$. Since $0$ is sent into $B_1(0)$, then $\mu(B_1(0))=B_1(0)$.
Exercise. Let $g:B_1(0)\to B_1(0)$ be a holomorphic function such that $g(\frac{1}{2})=0$. Then $|g(0)|\leq \frac{1}{2}$.
Proof. We'll use part of the proof of the Schwarz-Pick theorem, along with the facts that the Möbius transformations used are involutions.
Let $$\mu(z)=\frac{\frac{1}{2}-z}{1-\frac{z}{2}}.$$
Take $f=g\circ\mu$. Then $f(B_1(0))\subseteq B_1(0)$
and $f(0)=g\left(\frac{1}{2}\right)=0.$
Applying Schwarz Lemma, $|f(z)|\leq |z|$ for all $z$. But then,
$$\left|g\left(\frac{\frac{1}{2}-z}{1-\frac{z}{2}}\right)\right|\leq |z|,$$
which for $z=\frac{1}{2}$ gives
$$\left|g(0)\right|\leq \frac{1}{2},$$
the desired result.


Note: In the proof of the Schwarz-Pick Theorem the inverse $\mu^{-1}$ is used instead of $\mu$, and there is a second Möbius transformation which in this case is $\lambda(z)=-z$. So in this case $\mu=\mu^{-1}$ and $\lambda$ can be removed.

The Fundamental Group of a Sphere with Two Points Identified.

This is exercise 1.2.7. from Hatcher's book.
Lemma (Proposition 1.26.(a), Page 50 of Hatcher's Algebraic Topology).
If $Y$ is a cell complex obtained from a cell complex $X$ by attaching $2-$cells $e_{\alpha}^2$, $\alpha\in \Lambda$ by attaching maps $\phi_\alpha: S^1\to X$, then the inclusion $X\hookrightarrow Y$ induces a surjection $\pi_1(X,x_0)\to \pi_1(Y,x_0)$ whose kernel is $N=\langle[\gamma_\alpha \phi_\alpha\overline{\gamma}_\alpha]\mid \alpha\in \Lambda, \gamma_\alpha:x_0\to \phi_\alpha(1)\rangle$  where the $\phi_{\alpha}$ are seen as loops.
Exercise. Let $X$ be the quotient space of $S^2$ obtained by identifying the north and south poles to a single point. Put a cell complex structure on $X$ and use this to compute $\pi_1(X)$.

Solution. We can give $S^2$ the usual cell structure with one $0-$cell $a$ and one $2-$cell attached by the only map $S^2\to \{a\}$. Instead, we can give $S^2$ another cell structure from $S^1$ pasting two $2-$cells to the $1-$cell structure of $S^1$. We can also define a $2-$cell structure on $S^2$ by starting with two points $n,s$, two $1-$cells $\gamma,\delta$ between $n$ and $s$, and two $2-$cells pasted into $\alpha\beta$. This last approach works, since then we are giving $X$ the structure of the quotient complex $S^2/\{n,s\}$. It also gives a cell complex structure to $S^1$ starting with two points and attaching two $1-$cells.

We can instead think of it geometrically. When identifying the poles of the sphere with a point the resulting figure seems like a torus with a generating circle identified to a point. This determines the cell structure shown by the next drawing:

Here we have $X$ built from $S_1=\text{im}\alpha$ by attaching a $2$-cell by the map $\phi=\alpha\overline{\alpha}$ which is nullhomotopic. So, by the lemma, the homomorphism

$$\pi_1(S^1,v_1)\to \pi_1(X,v_1)$$

induced by the inclusion is an isomorphism, and $\pi_1(X,v_1)\cong \bZ$.