1

I am not getting the motivation behind defining Complex plane. I mean why defining such plane helps to understand $i$? So if I want, can I replace any other number in place of $i$( suppose, I replace it by $\Phi=\frac{\sqrt{5}-1}{2})$ to study properties of some number? Will that be helpful? If yes, why?

I have searched for this question, it is not asked before here. A similar question - Why the cartesian plane is defined like so?. But that is defferent from my question.

blub
  • 4,794
  • 2
    Maybe also relevant: https://math.stackexchange.com/questions/2325379/why-does-there-exist-an-imaginary-axis-on-the-argand-diagram, https://math.stackexchange.com/questions/1842968/why-is-the-complex-plane-shaped-like-it-is/1843044. – Hans Lundmark Aug 02 '18 at 08:41
  • @Ken You're completely entitled to make a new plane where the vertical axis is in units of $\frac{\sqrt5-1}{2}$, but this would be equivalent to a transformation of the normal complex plane. In other words, we could get your new plane simply by stretching the normal plane vertically by a factor of $\frac{\sqrt5-1}{2}$. So since it's fairly trivial what units we use, we may as well go for the neatest ones, don't you think? – Jam Aug 02 '18 at 08:45
  • We have a number line for the real numbers. I was first introduced to it when I learned addition, subtraction and multiplication, and it worked well. One can use it to leverage our visual intuition in order to better understand arithmetic.

    It seems natural to want to extend this success to the complex numbers.

    – Arthur Aug 02 '18 at 12:18
  • 1
    @Jam Note that $\frac{\sqrt{5}-1}{2}$ is a real number, so there are some problems with putting this on the vertical axis. – Servaes Aug 02 '18 at 12:29
  • @Servaes I thought it was clear that I meant $\ldots-\frac{\sqrt5-1}{2}i,0,\frac{\sqrt5-1}{2}i,2\frac{\sqrt5-1}{2}i,\ldots$ on the vertical axis – Jam Aug 02 '18 at 12:31
  • @Jam The body of the question talks of replacing $i$ by $\frac{\sqrt{5}-1}{2}$, which I conflated with your comment where you talk of vertical units of $\frac{\sqrt{5}-1}{2}$, which you then (somewhat confusingly) refer to as "your new plane". – Servaes Aug 02 '18 at 12:34

4 Answers4

5

The key idea is that complex arithmetic has a geometric interpretation: addition is translation, multiplication by real numbers is scaling, multiplication by $i$ is rotation by $90^{\circ}$. This is why it makes sense to think in geometrical terms, in terms of a plane.

Wouter
  • 7,673
4

Complex numbers are of the form $a+ib$ for real $a,b$ where $i$ is the imaginary constant. There are many good reasons as to why to define the complex plane and why to define it basically as the cartesian plane, but let me start with this:

You may identify a complex number $a+ib$ with a pair of real numbers $(a,b)$. The different operations you have may be turned into operations on tuples as well. You now transformed you concerns about complex numbers into concerns about objects in the real plane and may naturally plot complex numbers as points, where one axis now corresponds to the real part and one to the imaginary part.

Why is this useful? Well it turns out that complex numbers bear an intimate relationship with this representation. So instead of making things more ugly, you will find natural properties and natural geometric interpretations. For example addition turns out to be just vector addition in the plane and multiplication has a nice representation in terms of rotation and stretching.

This relationship is not only helpful in the realm of complex numbers but also the other way around. In processes called complexification, you can turn problems from real spaces into complex problems of lower dimension and find interesting properties allowing intriguing solutions back at the real case.

So, the definition of the complex plane follows from a natural identification of two objects. But once this correspondence is established, the use of it is justified through marvellous advancements/simplification on both sides

blub
  • 4,794
2

To unpick every aspect of your question would go quite deep. What you are asking has to do with field extensions, vector spaces, lattices, integers and geometry.

If you are looking at extensions of the Real Numbers, your $\Phi$ is already a real number, and adds nothing. But $\Phi$ is not a rational number, and generates a two-dimensional extension of the rationals, the integers of which can be treated as a lattice in a two-dimensional plane. If you had chosen the root of an irreducible cubic you would get three dimensions. Studying these extensions ends up being really fruitful (see Galois Theory)

$i$ has two special properties - first it is not a Real Number, and extending to the Complex Numbers gives an algebraically closed field - no further algebraic extensions are possible. Lattices in the complex plane are studied - the algebraic integers related to $\omega$ with $\omega^2+\omega+1=0$, for example, so that $\omega^3=1$.

The second is that geometrically, if we interpret $i$ as a rotation through $90^{\circ}$, everything works out beautifully and the points on the unit circle in the plane turn out to represent rotations through arbitrary angles. Multiplication and rotation are inextricably linked.

Neither of these really gets to the heart of "understanding $i$" - except that this particular number has some remarkably good properties. Other numbers have some of those properties - and these provide fruitful avenues of study and exploration. $i$ is special because it has them all.

And if we look at analytic properties rather than simply algebraic/geometric ones we find the Cauchy-Riemann equations and their remarkable consequences.

A huge amount of mathematics could be framed as "the consequences of the existence of $i$" - what we have inherited in the complex plane is just one window into that mathematics, which has been found to be a particularly useful point of view. You will not catch all aspects at first glance. It is worth holding on for the view.

You should also hold on to your intuition that $i$ shares properties with other numbers which might have to do with two dimensions. That is itself an interesting and fruitful avenue to explore.

Mark Bennet
  • 100,194
2

What you've suggested is perfectly fine, but as you've guessed, the usefulness is limited (geometrically). For example, if you take the rational numbers $\mathbb Q$, you can define "$j$" to be $\sqrt{2}$, and then you have a plane of points $$\mathbb{Q}(\sqrt{2}) = \{a+bj: a,b\in\mathbb{Q}\} = \{a+b\sqrt{2}: a,b\in\mathbb{Q}\}$$ You can already see that this looks very similar to the complex numbers - for the sake of this discussion, I'll call them the Qomplex numbers. But does it look like a plane in a natural way?

Let's see what multiplication looks like in $\mathbb{Q}(\sqrt{2})$: We have $$(a+bj)(c+dj) = ac + bcj + adj + bdj^2 = (2bd+ac)+(ad+bc)j.$$ It's a little bit like complex multiplication in the way it mixes up the coordinates. Geometrically, a number should have "length one" if it's Euclidean length is one. Suppose, in the same way was the complex numbers, we produce a vertical axis (the $\sqrt{2}\mathbb Q$ direction) and a horizontal axis (the $\mathbb Q$ direction). Then a Qomplex number $a+bj$ has "length 1" if $$\vert a+bj\vert^2 = a^2+b^2=1.$$ If we square a Qomplex number with length 1, it should remain length 1 - but we find that $$\vert(a+bj)(a+bj)\vert^2 = \vert a^2+2b^2 + 2abj\vert^2 = (a^2+2b^2)^2+(2ab)^2 = a^4 + 8 a^2 b^2 + 4 b^4.$$ Since $a^2+b^2$ doesn't divide this number, not only is the new length not one, we can't actually say anything about what it is (provided all we know is $a^2+b^2=1$).

If you pick $i = \sqrt{-1}$, then $|z||w| = |zw|$. This is exactly what we want - multiplying two numbers with length 1 should return a number with length 1.

If we're working in a plane, it's beautiful if the operations can correspond to simple movements in the plane. In the cartesian setting, vector addition is precisely what you want it to be - translating vectors by the amount you're adding to it. Here we have multiplication - what does this look like?

Let's take a Qomplex number $a+bj$ and multiply it by $j$. Then we obtain $2b+aj$. If $a+bj$ lives in the top right quadrant, so does the result. There is a strange mix of translation and rotation occurring. What about $i$? Well $i(a+bi) = -b+ai$. In the complex plane, this is precisely a rotation by $\pi/2$ and nothing more!

The complex plane is powerful and useful because everything behaves in beautiful geometric ways. Sure, we can define a "plane" such as the Qomplex numbers, but there is no geometric intuition. With the complex numbers, the operations simply match the geometry of a plane in the appropriate ways, and so we define the plane.

Harambe
  • 8,230
  • Thank you for your response, I am new to complex numbers, haven't we defined $|i|=1$? Please correct me if I'm wrong. –  Aug 02 '18 at 15:20