# Are imaginary numbers really necessary?

1. Dec 13, 2011

### nhmllr

You guys are probably sick of people who know little math posting here, but there's something that's been bugging me. I've bought The Feynman Lectures on Physics and have been reading through it slowly, and I'm up to the part where he talks about probability amplitudes of the electrons/photons. Now, I know that eix = cos x + i sin x. And when talking about adding up these probability amplitudes, the book often says things like "you have to add the real part of eix1 and the real part of eix2, then add the imaginary part of eix1 and the imaginary part of eix2, then square each and add them together to get the final probability." And the whole time, I think to myself, geez, why not just say add the horizontal components and the vertical components. Instead of saying, "the real part of eix" why not just say "cos x?" I get that Euler's Formula is pretty and that Feynman likes it (that much is on the very top of the wiki[/PLAIN] [Broken] article) but it seems to me that all this talk of imaginary numbers doesn't mean much. All they need is a 2-dimensional space to add up vectors, and they just so happened to choose the Argand Plane.

I'm not upset about it, but is there a real reason to use i, and not just some arbitrary Cartesian 2 dimensional space? Is it actually necessary that the unit of one of the components be the square root of negative one?

I'm just curious if this will pay off. Thanks.

Last edited by a moderator: May 5, 2017
2. Dec 13, 2011

### turbo

Yes, imaginary numbers are necessary and useful, especially in fields in which complex numbers are essential.

3. Dec 13, 2011

### jewbinson

Not all 2-d vector spaces are the same.

i = sqrt(-1), and so the interaction of (multiples of) i with the real numbers is different, interesting, and of importance in QM.

In most vector spaces we can have independent vectors which form a basis for that field, e.g. ( 0 1 ) and ( 1 0 ) is a basis for 1x2 matrices under scalar addition and multiplication of the reals, and is a 2-d v.s. .

We can never multiply ( 0 1 ) by a real number to get (a real multiple of - except 0) ( 1 0 ).

In fact, we can never multiply ( 0 1 ) by a complex number to get (a real multiple of - except 0) ( 1 0 ).

However, for any complex numbers z = x + iy, w = u + iv, we can always find a complex number b = c + id so that zb = w.

This useful property follows from the fact that i*i = -1, i.e. an imaginary number, seemingly different and unconnected to a real number, multiplied by another real number, gives us a real number!

4. Dec 13, 2011

### bp_psy

The complex space is not exactly some arbitrary 2D-vector space. It has a standard basis {(1,0),(0,1)} and inner product (a,b).(c,d)=ac-bd. You need this kind of object Qm and SR.Writing them down as a complex number z=a+bi is just a lot more convenient.

Last edited: Dec 14, 2011
5. Dec 13, 2011

### camilus

The complex plane is essentially R^2, but it has an extra condition that i^2=-1. This is the unique and unusual part, in nature, we don't encounter or count with numbers which we square and get a negative answer, they seem to be prevalent in the background picture however.

Imaginary numbers are one of the most fruitful and useful discoveries of mathematics, as well as most beautiful.

6. Dec 13, 2011

### turbo

Thank you!!

7. Dec 13, 2011

### camilus

It's my pleasure

8. Dec 13, 2011

### camilus

Sorry it's my cousin posting dumb ****. Disregard any weird nonmathematical posts

9. Dec 14, 2011

### PatrickPowers

You are largely right. Using a Cartesian 2D space would be essentially the same. The difference is that the complex notation is more compact. The operators you need are right there, and the i is useful when actually evaluating something. I would hate to do a Fourier transform with sines and cosines. It would be messy and obscure. In sum, complex numbers are neater and provide clarity once you get used to them.

Last edited by a moderator: May 5, 2017
10. Dec 15, 2011

### Chopin

You're right that complex numbers look a lot like vectors when you add them (technically speaking, their additive structure is isomorphic to $\mathbb{R}^2$). But they also possess another operation which is much different than a vector space: multiplication. Specifically, $Ae^{i\theta}Be^{i\phi}=ABe^{i(\theta + \phi)}$. This basically means that you can multiply a complex number by another complex number, and "rotate" the first through the complex plane by an angle $\theta$. This is not something that a normal vector space can do.

What's neat about this is that you can create a function $A(t)=Ae^{i\omega t}$, which describes the complex number $A$ rotating through the complex plane with frequency $\omega$. This has the very useful property that $\frac{d}{dt}A(t)=i\omega A(t)$, which becomes very important in lots of quantum mechanical equations.

11. Dec 15, 2011

### Jobrag

12. Dec 15, 2011

### AlephZero

They are not "necessary" but they are often very convenient. As other posts have said the reason is what happens when you multiply them. Just adding and subtracting complex numbers is not much different from using two real numbers, as you said.

It turns out that multiplying them is a neat way to do the math for many different things in physics and engineering, like analyzing electrical circuits, studying mechanical vibrations, aerodynamics, etc - not to mention quantum physics.

To be fair, you don't NEED complex numbers to do any of the above, but to keep track of the equivalent of complex multiplication you need to use FOUR real numbers not two, and two extra equations for each set of 4 numbers, so you have to do a lot more work. And at the end, the four numbers always come out a set of values that look like (a, b, -b, a), which should be warning flag that this isn't the best way to do the math.

13. Dec 15, 2011

### homeomorphic

You could say "necessary" is a matter of opinion. Maybe all math is unnecessary if you prefer to live primitively.

However, there is a sense in which complex numbers are necessary.

Usually, the lie they tell you is that imaginary numbers were invented so that any quadratic equation like

x2+ b x + c = 0

will have a solution.

Well, it's not exactly a lie. They were sort of invented for that, but it's more complicated than that. You would solve a quadratic equation using the quadratic formula. But sometimes it gives answers that doesn't make sense.

For example, x2+1 = 0.

If you use the quadratic formula on this, you get the square root of negative one. This, historically, was thought to be nonsense. However, the physician/mathematician Cardano sort of highlighted this and invented imaginary numbers. However, he then stated that his invention was "as subtle as it is useless".

So, basically, he just ended up saying that you could have imaginary numbers, but they are non-sense. Later on, Bombelli actually tried to multiply and add the things and work with them.

The thing that made people take imaginary numbers seriously was the solution of the cubic equation, one like the quadratic equation, but with a 3rd degree polynomial set equal to zero. There was a formula for this, but it wasn't completely understood for a while. Sometimes, like the quadratic formula, it gave square roots of negative numbers.

Bombelli noticed that, for certain a cubic equation, the formula gave square roots of negative numbers, but he knew the graph of the function crossed the x-axis so there had to be a real root. So, using the rules for multiplying and adding complex numbers, he found the root. And then, I guess, he must have checked that it was a solution by plugging it into the equation, and it worked.

So, the formula for the cubic involves complex numbers. Actually, there is a way around it. Viete came up with a way of solving it without complex numbers. But still, the success of complex numbers in solving a problem that you would think only concerns real numbers is a convincing argument that they ought to be taken seriously. This forshadows complex analysis, where sometimes you can use complex numbers to do ordinary real integrals that are otherwise hard to solve.

14. Dec 16, 2011

### nhmllr

I just want to say that I read everybody's posts. Thanks. :)