# What can complex numbers do that linear algebra cannot?

I think I have some good understanding of complex numbers. They are represented as a+ib where a,b are reals, and i² = -1. More precisely, multiplying a complex number by i causes a rotation by 90° in complex space, just like multiplying by a negative number causes a rotation by 180°.

Now there don't seem to be many use cases for complex numbers. Their defendants often mention electrical engineering, Fourier transform, physical signals, etc...

My question is, what makes complex numbers uniquely useful? For those specific applications, why not use vectors with 2 reals and do the rotations with cos and sin instead of i?

Since my question does not seem to be understood very well, let me provide more details.

For example, one thing that is original in complex numbers is multiplication. To multiply 2 numbers (a + ib) and (c + id) you actually end up with ac + ibc + iad + i²bd = (ac - bd) + i(ad + bc). However you could easily define this operation on R² as (a,b) x (c,d) = (ac - bd, ad + bc) and therefore (a,b)² = (a² - b², 2ab). From there, it is easy to redefine complex exponentiation, Euler's formula, etc... in term of basic 2D algebra using special, complex operations, just as special as dot or cross product.

If I go on, with all of complex operations, all of the results of complex numbers, what prevents me from getting rid of i and understanding everything in term of linear algebra?

• Complex numbers need defendants ? – Rene Schipperus Apr 27 '18 at 22:34
• I suggest you go to the complex-numbers or complex-analysis tags, and answer a few questions there using (only) linear algebra. Chances are that will put some things in perspective. – dxiv Apr 27 '18 at 22:37
• – angryavian Apr 27 '18 at 22:39
• Thanks, angryavian, that's the kind of stuff I was looking for. – d3m4nz3 Apr 27 '18 at 22:41
• They are an important subset of linear algebra - specifically, they let you more easily present things like Jordan normal form of matrices. You can come up with an equivalent formulation, but in some sense. you'd be just singling out the complex numbers as a specific type of matrix. – Thomas Andrews Apr 27 '18 at 22:56

After your edit, I can better understand your question.

Yes, if you define the multiplication as $$(a,b) \times (c,d) = (ac - bd, ad + bc)$$ $$\big($$and the addition as $$(a,b)+ (c,d) = (a+c, b+d)\big)$$

then you have just set up the algebraic structure of the complex field.

Since $$(0,1)\times(0,1)=(-1,0)$$

you can introduce the notation $$i=(0,1)$$

and you can simplify everything. From your point of view, you are right. But it took a few hundred years to see everything so clearly. You are standing on giants' shoulders.

The use that I was always told is that there are certain equations or problems which cannot be solved with the constraint of using only $x\in\Bbb R$, but that we can go via the complex plane to arrive at the solution.

• But C is equivalent to R², not R. Certainly, we can arrive to a solution for x² = -1, if we are replace x with a vector, and multiplication with rotation. – d3m4nz3 Apr 27 '18 at 23:01
• @demanze Chicken and egg. How might you go about computing the (generalized) real Jordan normal form of a real matrix that includes a rotation? – amd Apr 27 '18 at 23:34

In fact complex numbers, like negative numbers were not accepted for a long time. Gauss's proof of the fundamental theorem of algebra was decisive.