The series of functions in the complex plane. This is closely related to another question that I asked here regarding series of functions in the complex plane.
The $\sin(z)$ series in the complex plane is represented as $z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots$. In the real line with $\sin(x)$ we have Taylor's series and and the remainder and the convergence and such things. The question I have is---- how do we get this extension with the complex plane? In the question I have referred, people suggested that it was just a definiton, and that $\sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots$ was defined as such. But somehow that answer seems very unconvincing. It is like being told that the $\sin(x)$ series extension in the real plane is defined as such. It seems like it may have got something to  do with $\sin(z)$ extension in the complex plane having to satisfy the $\sin(x)$ component in the real line too (because real is a subset of complex) and since the coefficients in the real line is known to be $(1,0,\frac{1}{3},0,\frac{1}{5}......)$ the coefficients in the complex series expansion also have to be the same, although I dont know whether that line of approach is correct or whether it can be proven rigorously.
 A: Note: An extension of $\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$ from the real numbers to the complex numbers is not merely a matter of definition as some people seem to have suggested according to OP. It is the result of a sound concept based upon the uniqueness of analytic continuation of a series.

From the foreword of Complex Analysis by S. Lang:
More recent texts have emphasized connections with real analysis, which is important, but at the cost of exhibiting succinctly and clearly what is peculiar about complex analysis:
  
  
*
  
*the power series expansion,
  
*the uniqueness of analytic continuation, and
  
*the calculus of residues.

The uniqueness of analytic continuation is addressed in chapter II, section 3 by the following theorem:

Theorem 3.2
  
  
*
  
*(a) Let $f(T)=\sum a_nT^n$ be a non-constant power series, having a non-zero radius of convergence. If $f(0)=0$, then there exists $s>0$ such that $f(z)\ne 0$ for all $z$ with $|z|\leq s$, and $z\neq 0$.
  
*(b) Let $f(T)=\sum a_nT^n$ and $g(T)=\sum b_nT^n$ be two convergent power series. Suppose that $f(x)=g(x)$ for all points $x$ in an infinite set having $0$ as a point of accumulation. Then $f(T)=g(T)$, that is $a_n=b_n$ for all $n$.

Note that (a) tells us that $f$ is analytic in an open neighborhood of $0$. From (b) we conclude that unique extension is given if there is a small interval where $f$ and $g$ are equal. In fact equality only at an infinity of points with an accumulation point is already sufficient.
We can now use this theorem to uniquely extend real functions like $e^x, \sin(x)$ and $\cos(x)$. S. Lang continues with the following example:

Example: There exists at most one convergent power series $f(T)=\sum a_nT^n$ such that for some interval $[-\varepsilon,\varepsilon]$ we have $f(x)=e^x$ for all $x$ in $[-\varepsilon, \varepsilon]$. This proves the uniqueness of any power series extension of the exponential function to all complex numbers. Similarly, one has the uniqueness of the power series extending the sine and cosine functions.

This has nice consequences. Let us consider e.g.
\begin{align*}
\sum_{n=0}^\infty \frac{z^n}{n!}\qquad\qquad e^x(\cos y+i \sin y) \qquad \qquad\lim_{n\rightarrow \infty}\left(1+\frac{z}{n}\right)^n
\end{align*}
These expressions are analytic, and they all agree with $e^x$ when $z$ is real. We conclude according to Theorem 3.2 they must all be equal to each other, for they are all different expressions of the unique analytic continuation of $e^x$.
A: You're right about one thing: There are zillions of extensions of $\sin x$ into the complex plane, including some very smooth ones. For example,
$$f(x+iy) = e^y\sin x,\,\, x+iy \in \mathbb C.$$
The question is, of all these extensions, which one should we choose?
Forget about $\sin z$ for the moment and let me show you a few basic results in complex analysis. First is the idea of complex differentiability. If $f:\mathbb C\to \mathbb C,$ we say $f$ is complex differentiable at $z$ if
$$\lim_{h\to 0} \frac{f(z+h)-f(z)}{h}$$
exists. Here $h\to 0$ through complex values. We denote this limit by $f'(z),$ no surprise there. If $f'(z)$ exists for all $z\in \mathbb C,$ then $f$ is said to be entire (because it's complex differentiable in the entire complex plane).
If you haven't studied this idea yet, you're in for a big surpise (quite a pleasant one) when you do. First shocker: If $f$ is entire, then $f$ has complex derivatives of all orders everywhere. Second shocker: If $f$ is entire, then $f$ is everywhere equal to its Taylor series based at any point you like. Third shocker: If $f,g$ are two entire functions and $f=g$ on $\mathbb R,$ then $f=g$ on all of $\mathbb C.$ Clearly these entire functions are amazing creatures.
We've seen that entire functions equal everywhere-convergent power series. The converse is also true: If $ \sum_{n=0}^{\infty} a_nz^n$ converges for all $z\in \mathbb C,$ then the power series defines an entire function. The converse shouldn't be surprising at all; it is very much like results you've seen on $\mathbb R.$
OK, back to $\sin x.$ Through geometry and calculus, we prove
$$\sin x = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1)!}x^{2n+1}$$
for all $x\in \mathbb R.$ But this power series converges if $x$ is replaced by any $z\in \mathbb C$! Thus we have an entire function $f$ defined by
$$f(z) = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}z^{2n+1}$$
such that $f(x)=\sin x$ for $x\in\mathbb R.$ By the remarks above, it is the only entire function with this property. I would say the case for defining $\sin z$ with this power series is pretty darned good! 
A: I think that the approach you're looking for is the following. You know that $\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots$ (note the factorials, this is the right sine series). Now, to define $\sin$ in the complex plane you need to extend this real definition properly. There are two ways to do this:
1st: Define $\sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots$. Of course you need to prove that this makes sense, but the convergence argument follows from the real one case so ok. From this setting the answer to your question becomes "by definition".
2nd: Define $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$. From this setting you can prove that
$$
\sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots \ \ \ \ (\star)$$
as follows; the exponential functions are entire so by this definition $\sin(z)$ also is. Call $S(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots$. The real argument for the convergence of the sine series tells you that $S$ is an entire function. From here that $Z:=\sin-S$ is entire too. Furthermore you know that $(\star)$ holds for real $z$, so the entire function $Z$ vanishes in $\mathbb{R}$. By elementary complex analysis the zeros of holomorphic functions are isolated unless the function is the identical zero function, so it must be $\sin=S$; i.e. $(\star)$ hold for every $z\in\mathbb{C}$.
Notice that the technique about the zeros allows us to extend identities such as $\sin(x+y)=\cdots$ to the complex context. And note also that it can be argued which one of both definitions is more natural, or that the "source of inspiration" for the second definition is exactly the first one...
