Why does the imaginary number $i$ satisfy $i\times 0=0$? Why does the imaginary number $i$ satisfy $i\times 0=0$?
I mean, we don't really know what $i$ is. How could we be sure about that? I think there's a reason behind why mathematicians decided that. 

Ignore this answer if you've never heard of matrix multiplication. Better yet, learn matrix multiplication and then read this answer!
The answer to your question depends on what definition of complex numbers you're using. 
For example, I like to think of the complex numbers as matrices of the form
$$
\begin{pmatrix}
a & -b \\ b & a
\end{pmatrix}
$$
where $a$ and $b$ are real numbers. This allows us to define two complex numbers
\begin{align*}
\mathbf 0
&=
\begin{pmatrix}
0 & 0\\ 0 & 0
\end{pmatrix} &
i
&=
\begin{pmatrix}
0 & -1 \\ 1 & 0
\end{pmatrix}
\end{align*}
Hence we have the identity
$$
\mathbf 0\cdot i
= 
\begin{pmatrix}
0 & 0\\ 0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & -1 \\ 1 & 0
\end{pmatrix}
=
\begin{pmatrix}
0\cdot 0+0\cdot 1 & 0\cdot(-1)+0\cdot 0\\
0\cdot 0+0\cdot 1 & 0\cdot (-1)+0\cdot 0
\end{pmatrix}
=
\begin{pmatrix}
0 & 0\\ 0 & 0
\end{pmatrix}
=
\mathbf0
$$
One of the nice things about defining the complex numbers this way is that we avoid the confusing equation $i=\sqrt{-1}$. We also don't have to resort to using silly words like "imaginary."
Note, however, that we do have
$$
i^2
=
\begin{pmatrix}
0 & -1 \\ 1 & 0
\end{pmatrix}^2
=
\begin{pmatrix}
0 & -1 \\ 1 & 0
\end{pmatrix}\begin{pmatrix}
0 & -1 \\ 1 & 0
\end{pmatrix}
=
\begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix}
$$
So, if we define the complex number
$$
\mathbf{1}=
\begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix}
$$
then we recover the formula
$$
i^2=-\mathbf{1}
$$
To convince yourself that our definition of the complex number $\mathbf1$ is not arbitrary, note that $\mathbf 1$ enjoys the property
$$
\mathbf 1
\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
=
\begin{pmatrix}
1 & 0\\ 0 & 1
\end{pmatrix}
\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
=
\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
$$
This is remarkably similar to the celebrated equation $1\cdot a=a$. It's also worth noting that the complex number $\mathbf0$ satisfies
$$
\mathbf0+
\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
=
\begin{pmatrix}
0&0\\0&0
\end{pmatrix}+
\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
=
\begin{pmatrix}
0+a & 0-b \\
0+b & 0+a
\end{pmatrix}
=
\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
$$
which is similar to our usual equation $0+a=a$.
The point here is that complex-arithmetic "feels" like ordinary arithmetic but is indeed different. As @Hurkyl points out, algebraic gadgets that behave like this are called rings.
