What is complex multiplication?

When visiting the wood worm, I saw a very old math post being dug up, this post is probably the discussion if the plural is regarded as a vector, then how should plural multiplication be treated? Are there multiplication between vectors? For example, the plural $ (1+i) $ and the plural $i$, their corresponding vectors are $\left[{\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]$ and $\left[{\begin{array}{*{20} {C}} 0\\ 1 \end{array}} \right]$, then two vectors how to get the complex $-1+i$ corresponding vector $\left[{\begin{array}{*{20}{c}} -1\\ 1 \end{array}} \right Where's]$?

In fact, I think it's not enough to treat a plural as a vector! Now that we think of the plural as vectors, we should also discuss (linear) transformations , which is made sense. So, if we think of vectors as linear transformations, then the result is trivial!

We know that each non-0 complex number has an exponential form (exponential form): $z = R{e^{i\theta}}$. The ${e^{i\theta}}$ is a "rotation transformation", that is, a Shang clockwise rotation $theta$ angle, in the perspective of linear algebra, its corresponding linear transformation is $\left[{\begin{array}{*{20}{c}}
{\cos \theta}&{-\sin \theta}\\
{\sin \theta}&{\cos \theta}
\end{array}} \right].$ Therefore, each complex number can uniquely correspond to a linear transformation: $ $z = R{e^{i\theta}} \sim r\left[{\begin{array}{*{20}{c}}
{\cos \theta}&{-\sin \theta}\\
{\sin \theta}&{\cos \theta}
\end{array}} \right].$$ so, complex multiplication $z_1 * z_2$ We can think of the plural $z_1$ as a linear transformation $t$, and the plural $z_2$ as its corresponding vector $v$, there are $ $z _1 * z_2 \sim T v.$$ Finally we put the result $T v$ (a vector) corresponds to the number of reply fields.

Example. Computes complex multiplication by $ (1+i) *i$.

First, it's easy to know that $ + i = \sqrt 2 {e^{i\frac{\pi}{4}}} \sim \sqrt 2 \left[{\begin{array}{*{20}{c}}
{\cos \frac{\pi}{4}}&{-\sin \frac{\pi}{4}}\\
{\sin \frac{\pi}{4}}&{\cos \frac{\pi}{4}}
\end{array}} \right] = \left[{\begin{array}{*{20}{c}}
1&{-1}\\
1&1
\end{array}} \right]$, and $i \sim \left[{\begin{array}{*{20}{c}}
0\\
1
\end{array}} \right]$, then $\left[{\begin{array}{*{20}{c}}
1&{-1}\\
1&1
\end{array}} \right]\left[{\begin{array}{*{20}{c}}
0\\
1
\end{array}} \right] = \left[{\begin{array}{*{20}{c}}
{-1}\\
1
\end{array}} \right] \sim-1 + i$. So $\left ({1 + i} \right) i =-1 + i$.

What is complex multiplication?