I haven’t taken any complex analysis course yet, but now I have this question that relates to it.

Let’s have a look at a very simple example. Suppose x,y and z are the Cartesian coordinates and we have a function z=f(x,y)=cos(x)+sin(y). However, now I change the R2 plane x,y to complex plane and make a new function, z=cos(t)+isin(t).

So, can anyone tell me some famous and fundamental differences between complex plane and R2 by this example, like some features R2 has but complex plane doesn’t or the other way around. (Actually I am trying to understand why electrical engineers always want to put signal into the complex numbers rather than R2, if a signal is affected by 2 components)

Thanks for help me out!

**Answer**

R2 and C have the same cardinality, so there are (lots of) bijective maps from one to the other. In fact, there is one (or perhaps a few) that you might call “obvious” or “natural” bijections, e.g. (a,b)↦a+bi. This is more than just a bijection:

- R2 and C are also metric spaces (under the ‘obvious’ metrics), and this bijection is an isometry, so these spaces “look the same”.
- R2 and C are also groups under addition, and this bijection is a group homomorphism, so these spaces “have the same addition”.
- R is a subfield of C in a natural way, so we can consider C as an R-vector space, where it becomes isomorphic to R2 (this is more or less the same statement as above).

Here are some differences:

- Viewing R as a ring, R2 is actually a direct (Cartesian) product of R with itself. Direct products of rings
*in general*come with a natural “product” multiplication, (u,v)⋅(x,y)=(ux,vy), and it is not usually the case that (u,v)⋅(x,y)=(ux−vy,uy+vx) makes sense or is interesting in general direct products of rings. The fact that it makes R2 look like C (in a way that preserves addition and the metric) is in some sense an accident. (Compare Z[√3] and Z2 in the same way.) - Differentiable functions C→C are not the same as differentiable functions R2→R2. (The meaning of “differentiable” changes in a meaningful way with the base field. See complex analysis.) The same is true of linear functions. (The map (a,b)↦(a,−b), or z↦¯z, is R-linear but not C-linear.)

**Attribution***Source : Link , Question Author : Cancan , Answer Author : Billy*