See, math, it's dangerous (maybe more than I thought). Oh well, this title should be sufficient. If anything, it thematically connects this post to my last post's brief mention of how complex analysis is awesome for geometry, symmetry, and pretty much everything.
Last post I gave a brief introduction to symmetry. I figured I could take the time and make another post to detail the basics of complex analysis. Again, this will be long and will simply give an introduction to the use of complex numbers, so feel free to skip through it if you know it already.. Soon, I'll make another post for group theory.
Complex numbers are awesome. They are very similar to the real numbers, but by the way we construct them, allow for greater geometric significance. First let's begin with i.
i is the square root of -1, defined by the equation i2 = -1. This allows us to define complex numbers, where we can more fully account for algebra. For example, when we are restricted to just the reals, the equation x4 - 1 = 0 has just two solutions--1 and -1--but in the complex numbers it has 4 (the same amount as the degree of the polynomial)--1, -1, i, and -i.
Now let's define complex numbers
Take R2
No, not you |
Excuse me, the real plane, R2
It has an x-axis and a y-axis, and every point in R2 can be represented by a coordinate pair, (a, b).
The complex plane is very similar, instead of an x- and a y-axis, it has a "real" and an "imaginary" axis:
Here, still every point can be represented (a, b), if we understand (a, b) --> a + ib. This is the form of a complex number--a is the real part, and ib is the imaginary part.
Every complex number, as I alluded to in my last post, has an argument and a modulus. The argument is the angle, and the modulus is the distance between (a, b) and the origin, (0, 0). Thus, the modulus of z = a + ib is |z|2 = a2 + b2.
Complex numbers can also be represented by z = r cis(θ)= r(cos(θ)+ isin(θ)) = r eiθ, where r is a real number and represents the modulus, and 0≤ θ≤ 2π where θ represents the argument.
The usual algebraic operations work with complex numbers, and it is easiest to view i as a variable with the special case that i2 = -1.
Let w = a + ib and v = c + id be complex numbers. Addition and subtraction are clear: w+ v = (a+b) + i(b+ d). Multiplication, less so, but still intuitive: wv = (a + ib)(c + id) = ac + iad+ ibc - bd = (ac - bd) + i(ad + bc).
Now that we see multiplication, let me introduce you to our friend the complex conjugate! Each complex number z = a + ib has a complex conjugate = a - ib. Notice how this is the reflection of z across the real axis!
From wikipedia |
Now, this gives us motivation behind division. v != 0, w/v = (a+ ib)/(c+ id), and multiplying by the conjugate of v over the conjugate of v, gives us ((ac+ db)+ i(bc - ad)) / c2 + d2.
In polar coordinates, these operations becomes much smoother. Multiplication of w= r1 eiθ1 and v = r2 eiθ2, wv= (r1r2) ei(θ1+ θ2). Notice that the multiplication of the conjugate would result in some z = (r r ) ei(θ- θ = r2 e0 = r2, which is exactly what we'd think it'd be.
Inversion is easy to see. We are looking for a z-1 such that zz-1 = 1. If z = r eiθ then by what we just saw before, we are looking for some z-1= p eiΦ such that rp = 1 and θ+ Φ= 0. Thus, p = 1/r and Φ= -θ.
This post got pretty long, so I'll give you a break and will post about mapping in the complex plane and the Riemmanian sphere in a separate post later.
Connor
So I definitely lost my train of thought about half way through your explanation. Have you ever considered making a video to explain a math concept? Something like Khan Academy? I'm not sure how much work it would take to put one together, but it might be a cool addition to your blog! If not, your written explanations are quite thorough and interesting so no worries :)
ReplyDeleteps I really appreciated the R2D2
Hey Connor, this was a great intro post on Complex Numbers. I'm excited to see how you apply the Complex Numbers to the Mobius Transformations!
ReplyDeleteNice post. In one place you wrote cis for cos but who's counting?
ReplyDeleteExcuse me that this wasn't clear, but cis(x) = cos(x)+ isin(x), which is equal to e^(ix), as can be shown with Taylor series approximations. That's where we get the famous expression e^(i*pi) + 1 = 0. Cos(pi) + isin(pi) = -1 + 0.
DeleteDoes every imaginary number have a conjugate?
ReplyDeleteA conjugate is just flipping the sign in the imaginary portion of a complex number, so yes, every imaginary number has a conjugate.
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