# Complex plane

1. Aug 13, 2014

### DiracPool

I'm trying to get straight some basic complex number fundamentals.

First we have z=x+iy. Ok, my question is what does the z in the equation represent? Does it represent a "point" on the complex plane? Does it represent a vector from the origin to where the x and iy values add? Does it represent the "length" of that resultant vector?

I'm guessing it doesn't represent the length of the vector, z, because I read that the length of that vector is the squared modulus of z^2=x^2+(iy)^2. I'm guessing that if we didn't take the modulus, we'd end up with an equation which gave us a length of z=√x^2-y^2. Why are we allowed just to take the modulus here and not worry about it? Just because it gives us a Pythagorean "real" looking length? It seems incorrect. It seems as though the length should be as the equation states, z=√x^2-y^2. What am I missing here?

But my main question is what does the z in the z=x+iy represent? And could you give me the equivalent on how that would look on just the ordinary "real" x-y plane.

I know this sounds elementary, but I just haven't been able to find a straightforward explanation anywhere.

2. Aug 13, 2014

### disregardthat

z can represent many things. The most ordinary way to imagine it is as a point in the complex plane with coordinatex (x,y). By the way, the modulus is |z|=√(x^2+y^2). But most of the time you will have to think of it as a number which can be added, multiplied, used as function arguments and values etc..

3. Aug 13, 2014

### MrAnchovy

You have this the wrong way round. A complex number $z$ is defined by the sum of a real part $x$ and the product of an imaginary part $y$ and the constant $i$ where $i^2 = -1$.

If we have two such numbers $z = a + ib, w = c + id$ we can add them together: $z + w = a + ib + c + id = (a+c) + i(b+d)$ or multiply them $zw = (a + ib)(c + id) = ac +ibc + iad + i^2bd = (ac-bd) + i(ad+bc)$; anything we can do with complex numbers can be done without reference to the complex plane.

If we want, we can represent a complex number by a point on the complex plane. And we can represent a point on a plane by a vector from the origin. But these are still just representations of the complex number z, they don't define it.

As for the modulus, this is simply the length of the vector representing z, $\lvert z \rvert = \sqrt{x^2 + y^2}$.

The complex plane is just an imaginary (!) construct where we take an ordinary plane with real coordinates (is this what you mean by a "real plane"?) and define the y coordinate to represent the coefficient of i.

Last edited: Aug 13, 2014
4. Aug 13, 2014

### HallsofIvy

Numbers, whether real or imaginary, "represent" whatever we want them to represent. Yes, we can think of the complex number, a+ bi, as "representing" ("represented by") a point in the complex plane. And we can represent a vector in the plane by either a pair of numbers (a, b), or the complex number, a+ bi. I would be cautious about either of those "representations" because vectors have operations defined that points do not and complex numbers have operations defined that vectors do not.

You mis-read. The modulus of a complex number, z= x+ iy, is given by $\sqrt{z\overline{z}}= \sqrt{(x+ iy)(x- iy)}= \sqrt{x^2+ y^2}$

And the length of vector represented by z (or the distance from the origin to the point represented by z) is the modulus, not the "square of the modulus".

5. Aug 13, 2014

### Hertz

Don't over-complicate it man. A complex number $z=x+iy$ is exactly that, a number. It's a number which has two components in the sense that you must specify $x$ AND $y$ in order to fully define the number $z$.

To specifically answer your question: Yes, a complex number is a point in the complex plane. Just like you can find any real number by moving side to side on a number line, you can find any complex number by moving side to side and up and down in the complex plane. A real number has a single numerical value which defines it, which is why you need only one dimension to represent the set of real numbers. However, a complex number has two values which define it, which is why you must use a plane to represent the set of complex numbers.

Even though the motivation behind needing such a set of numbers can be mysterious, you shouldn't over-complicate the basic idea. The complex numbers are just a set of numbers that are '2 dimensional' in that they take 2 real numbers to define (not sure if dimension is used right here). A complex number shares many properties of a 2D vector, but then again there are certain properties of vectors which complex numbers do not share, such as dot products and cross products and gradient operations and stuff like that. You surely can't add the components of a complex number to the components of a vector, so I wouldn't recommend considering a complex number to be a vector. Just consider it to be a number and as you explore properties of these types of numbers you will see that they share many properties with 2D vectors.

6. Aug 13, 2014

### DiracPool

Thanks to everyone who replied, I think I understand it better now. Z is just a point in the complex plane, and the length or modulus of the vector represented by z is found by multiplying the expression z=x+iy by its complex conjugate and then taking the square root. Is that correct?

7. Aug 13, 2014

### micromass

That is correct.

8. Aug 14, 2014

### haruspex

Sort of, but I would add my voice to the cautions expressed elsewhere in this thread with regard to confusing complex numbers with 2D vectors.

The complex field can be mapped to the points of a two dimensional Euclidean plane. This helps with visualising complex numbers and the operations they can undergo. It does not mean a complex number is a point in a plane.

A two dimensional vector space over the real field, $\Re$, can also be mapped to the points of a two dimensional Euclidean plane, with similar benefits.

There are similarities between the mappings. In each case the magnitude of the entity mapped (as defined separately for the two contexts) corresponds to the distance from the origin to the mapped point. The addition operations also work the same way in the mapping: if complex numbers w, z map to points A, B, and vectors X, Y also map to A, B, then map(z+w) = map(X+Y). But note that the two addition operations are essentially different: one operates on complex numbers while the other operates on vectors.

The difference becomes clearer when we look at multiplication. There's no multiplication operator defined on 2D vectors which produces a vector result, and none defined on complex numbers that guarantees to produce a real scalar result.
Further, consider that a 2D vector space could be defined over the complex field, so each vector is itself represented by two complex co-ordinates. (The scalar product gets tricky here. If X, Y are vectors then the scalar product X.Y is a complex number, and it is the conjugate of Y.X.)

9. Aug 14, 2014

### micromass

I see your point, but I don't think there's much wrong with thinking that a complex number is a point in the plane. For example, many constructions of $\mathbb{C}$ will define $\mathbb{C} = \mathbb{R}^2$ with some operations. In that sense, we have defined a point in $\mathbb{C}$ literally as a point in the plane.

10. Aug 14, 2014

### haruspex

No, it's a mapping, not an equality. Part of the definition of the complex numbers, ℂ, is that they have a dyadic operation (multiplication) defined on them with certain properties. That is not part of the definition of $\mathbb{R}^2$. Hence the two are different mathematical entities.

Last edited: Aug 14, 2014
11. Aug 14, 2014

### micromass

In my opinion that's a way too formal way to look at things. In the same way you should then argue that $\mathbb{R}\subseteq \mathbb{C}$ is not true.

12. Aug 14, 2014

### disregardthat

I tend to agree with micromass here. What do we really mean when we say $\mathbb{R}^2=\mathbb{C}$ with some operations? Surely, the tuple set theoretical construction $\mathbb{C} "=" (\mathbb{C},+,*,0)$ (or whatever, depending on context) is implicit in some sense when interpreting it rigorously. Just as a topological space constructed as $X = (\mathscr{T},\overline{X})$ where $\overline{X}$ is a set and $\mathscr{T}$ a topology on $\overline{X}$, we do say both that X is a topological space AND X has points $P \in X$. But, as a set theoretical pair, only $\mathscr{T}$ and $\{\mathscr{T},\overline{X}\}$ are set-theoretically contained in $X$.

In addition, you would have to work with a very strange definition of $\mathbb{R}$ to have $\mathbb{R} \subseteq \mathbb{C}$ set-theoretically as well.

Much of the underlying set theory in mathematics is implicit. All the time we give "new meaning" to $\in$ and $\subseteq$ when introducing new things such as groups, rings and topological spaces. If seen as a problem, this all lends itself towards a topos-like foundation of mathematics, where these things are given sense in terms of morphisms with particular properties (which does not in fact require a constant redefinition of terms).

Last edited: Aug 14, 2014
13. Aug 14, 2014

### haruspex

I understand I'm being pedantic, but with good cause. It is even more reasonable to equate $\mathbb{R}^2$ with a two dimensional vector space over $\mathbb{R}$. But combining those equations leads to the confusion that $\mathbb{C}$ is that vector space, with consequent pitfalls regarding multiplication etc. Most students cope with this double identity, but some do get tangled up. Better to have the mapping formalism spelled out once.

14. Aug 14, 2014

### disregardthat

To equate $\mathbb{R}^2$ with a two-dimensional vector space over $\mathbb{R}$ is also wrong in many contexts for the same reasons you have given. As topological spaces then, they are not equal, and not as rings, not as abelian groups etc..

15. Aug 14, 2014

### micromass

Notations like $\mathbb{R}$ and $\mathbb{C}$ are incredibly overloaded. It's usually no problem though, but I guess that newcomers might find it confusing. For example, $\mathbb{R}$ can be used to denote the set, the field, the topological space, the algebra, the vecotr space, and so on. Same with $\mathbb{C}$.

We interpret $\mathbb{C}$ as set all the time when we say $z\in \mathbb{C}$. And in these contexts, I find it totally reasonable to say $\mathbb{R}^2 = \mathbb{C}$. Of course, $\mathbb{C}$ has other structures that we tend to ignore some times. Also, I don't think it's wrong to see $\mathbb{C}$ as an $\mathbb{R}$-vector space.

I remember being in a talk by undergrads once, and some girl said "As we all know, the topological space $\mathbb{R}$ is locally compact". At the end of the talk, a professor said that she should always state which topology she was working with. I thought that was a pretty incorrect comment as such things are clear from context. Anyway, some professors disagreed with him too and started an entire discussion about it.

16. Aug 14, 2014

### disregardthat

Wasn't there a story about a student at an exam answering "$\mathbb{R}$" to the question "name a compact space". The professor, giving the student the benefit of the doubt, then asked which topology the student had in mind.

17. Aug 14, 2014

### micromass

Haha yeah. That one featured once as one of the "jokes that 99% of the population won't understand"

18. Aug 14, 2014

### haruspex

I agree. My point was that equating $\mathbb{R}^2$ with a 2D space over $\mathbb{R}$ is at least as defensible, if not more so, than equating it with $\mathbb{C}$; and doing both leads to confusion.