Is $\mathbb{C}^m\times\mathbb{C}^n=\mathbb{C}^{m+n}$ true? In real analysis we have
$\mathbb{R}^{m} \times \mathbb{R}^{n} = \mathbb{R}^{m+n}$, e.g.
$$
\mathbb{R}\times\mathbb{R}=\mathbb{R}^2=\big \{
(x,y): x\in \mathbb{R}, y\in\mathbb{R}
\big\}
$$
Is this also true for complex numbers, i.e. $\mathbb{C}^m\times\mathbb{C}^n=\mathbb{C}^{m+n}$?
 A: As in, the Cartesian product of sets? In order to answer that you have to recall the recursive definition:
$$X^1:= X$$
$$X^n:=X\times X^{n-1}$$
(there are alternative yet equivalent ways to define $X^n$) Now since we have the natural bijection
$$A\times (B\times C)\to (A\times B)\times C$$
$$(a, (b,c))\mapsto ((a,b),c)$$
it follows that the Cartesian product is "associative". With that property you can easily prove that
$$X^{n+m}\simeq X^{n}\times X^{m}$$
where $\simeq$ stands for "is equinumerous to". You do that by simply "rearranging brackets" (composing multiple natural bijections defined above) on the right side to match $X\times (X\times (X\times\cdots X)\cdots )$ pattern.
This bijection is good. As in really good. So good that in almost every scenario you encounter it will also preserve additional structures, e.g. if $X$ is a group then $\simeq$ is an isomorphism of groups. If $X$ is a topological space then $\simeq$ is a homeomorphism, etc. It is so good that mathematicians simply say that
$$X^{n+m}=X^n\times X^m$$
even though this is formally incorrect.
All in all, yes $\mathbb{C}^{n+m}=\mathbb{C}^n\times \mathbb{C}^m$.
A: Formally, the answer is negative: $\mathbb{C}^m\times\mathbb{C}^n$ is a set of ordered pairs, whereas $\mathbb{C}^{m+n}$ is a set of $(m+n)$-plets; therefore, they cannot possibly be equal, unless $m=n=1$ (in which case, yes, they are equal).
However, there is a natural bijection $B\colon\mathbb{C}^m\times\mathbb{C}^n\longrightarrow\mathbb{C}^{m+n}$, given by$$B\bigl((z_1,\ldots,z_m),(w_1,\ldots,w_m)\bigr)=(z_1,\ldots,z_n,w_1,\ldots,w_m),$$which allows us to identify these spaces.
Of course, there is nothing pecular about the complex numbers heere. The same identification can be done with any set $A$ instead of $\mathbb C$ (and, in particular, with $\mathbb R$).
