Why is $\frac{\sin(z)}{z}$ analytic at $z=0$ I am attempting exercise $11.2.9 a$ from Mathematical Methods for Physicists.
Having a look at a provided solution it states that:


*

*The derivative is: $f'(z) = \frac{\cos(z)}{z}-\frac{\sin(z)}{z^2}$. How can this be derived? I know that the quotient rule was used to determine the derivative but must one not prove that the function is analytic before using standard derivation tools? Or can one say that the function is analytic because it is only expressed in terms of $z$?

*The function is analytic everywhere except at infinity. I am having a hard to seeing this as I thought it is analytic everywhere except at $z=0$. 

 A: It's not the function $\frac{\sin z}{z}$ that's analytic over $\mathbb{C}$, but the function
$$
\operatorname{sinc}{z}=
\begin{cases}
\dfrac{\sin z}{z} & z\ne0 \\[6px]
1 & z=0
\end{cases}
$$
This is the case because
$$
\sin z=z\biggl(\,\sum_{k\ge0}\frac{(-1)^{k}z^{2k}}{(2k+1)!}\biggr)
$$
so we obtain the power series
$$
\operatorname{sinc}z=\sum_{k\ge0}\frac{(-1)^{k}z^{2k}}{(2k+1)!}
$$
which makes it analytic.
In general, every time a function $f$ is analytic over a domain $\Omega$ (connected open set) containing $z_0$, also the function
$$
g(z)=
\begin{cases}
\dfrac{g(z)-g(z_0)}{z-z_0} & z\ne z_0, z\in\Omega \\[6px]
g'(z_0) & z=z_0
\end{cases}
$$
is analytic over $\Omega$.
A: Let's collect the first building blocks for intuition. A clear view of math on the real number line is a foundation for a clear view of math in the complex plan. 
Because you can redefine the function as
$$
\frac{\sin z}{z} = 
\begin{cases}
 \frac{\sin z}{z} & x\ne 0 \\
 1 & x = 0
\end{cases}
$$
The problem is at the point $z=0$. Notice that
$$
\lim_{x\to0^{+}} \frac{\sin x}{x} = \lim_{x\to0^{-}} \frac{\sin x}{x} = 1.
$$
The right and left limits agree, we only need to correct the definition of the function. As pointed out by @mrf 3, the limits from any direction will converge to unity.
As noted by @Michael Burr, this is not a full answer to the question, but a tunnel through one obstacle to understanding.
A: You know $\sin z$ is analytic everywhere, so rewrite $\sin z$ in terms of its power series expansion
$$
\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots
$$
Then $\frac{\sin z}{z}$ is 
$$
\frac{\sin z}{z}=1-\frac{z^2}{3!}+\frac{z^4}{5!}+\cdots
$$
which you have just represented as a power series valid at zero. Thus that the given function is analytic at $0$ can be seen directly.
Also note, computing it's derivative is now easy, just differentiate termwise and plug in $z=0$!
A: First, observe that $\sin(z)$ is an entire function and 
$$
\sin(z)=\sum_{n=0}^\infty (-1)^n\frac{z^{2n+1}}{(2n+1)!}.
$$
Away from zero, we know that 
$$
\frac{\sin(z)}{z}=\sum_{n=0}^\infty (-1)^n\frac{z^{2n}}{(2n+1)!}.
$$
Even though $\frac{\sin(z)}{z}$ appears to be undefined at $z=0$, there is an analytic continuation given by the power series expansion: 


*

*The power series expansion is defined at $z=0$.

*The power series expansion agrees with $\frac{\sin(z)}{z}$ away from zero.

*The radius of convergence of the power series agrees with the radius of convergence for $\sin(z)$.
Since $\sin(z)$ and $z$ are entire, the only possible problems for $\frac{\sin(z)}{z}$ are at $z=0$ and $z=\infty$.  We've seen above that the function is analytic at $z=0$, so the only possible problem is at $z=\infty$.
We can see that the function is not analytic at $z=\infty$ by showing that it's not continuous at $z=\infty$.  In particular, as if we write $z=a+bi$ and look at $a\rightarrow\infty$ and $b=0$, we get that the function approaches zero (the numerator is bounded).  On the other hand, if you consider $a=0$ and $b\rightarrow\infty$, observe that $\sin(bi)=-i\sinh(b)$ and the limit $\frac{-i\sinh(b)}{bi}$ as $b\rightarrow\infty$ diverges to infinity.
A: First, the function needs to be defined at zero, and it is straightforward to establish using L'Hôpital's that $\lim_{z \to 0} \dfrac{\sin z}{z} = 1$. Denote the resulting function by $\operatorname{sinc}$.
Since $z \mapsto z$ and $\sin$ are entire, it follows that $ \operatorname{sinc} $ is analytic for $z \neq 0$, so the only question is for $z=0$.
To check the existence of the derivative of $\operatorname{sinc}$ 
at $z=0$, we look at $\lim_{z \to 0} \dfrac{\operatorname{sinc} z - \operatorname{sinc} 0}{ z}  $.
The power series for $\sin$ can be written as
$\sin z = z + z^3 g(z)$, where $g$ is analytic at $0$, hence
we have $\dfrac{\operatorname{sinc} z - \operatorname{sinc} 0}{ z} = \dfrac{\sin z -z }{ z^2} = z g(z)$, from which it follows that
$\lim_{z \to 0} \dfrac{\operatorname{sinc} z - \operatorname{sinc} 0 }{ z} =0$.
Hence $\operatorname{sinc}$ is entire.
We say that $f$ is analytic at $z=\infty$ if the function $z \mapsto f(\dfrac{1 }{ z})$ is analytic at $z=0$.
Note that for real, non zero $x$, we have $\operatorname{sinc} {1 \over ix} = -x \sinh (-\dfrac{1 }{ x})$, and so we see that $z \mapsto \operatorname{sinc} \dfrac{1 }{ z}$ is unbounded near $z=0$ hence not analytic.
