It's always taken for granted that the real number line is perpendicular to multiples of $i$, but why is that? Why isn't $i$ just at some non-90 degree angle to the real number line? Could someone please explain the logic or rationale behind this? It seems self-apparent to me, but I cannot actually see why it is.

Furthermore, why is the real number line even straight? Why does it not bend or curve? I suppose arbitrarily it might be strange to bend it, but why couldn't it bend at 0? Is there a proof showing why?

Of course, these things seem natural to me and make sense, but why does the complex plane have its shape? Is there a detailed proof showing precisely why, or is it just an arbitrary choice some person made many years ago that we choose to accept because it makes sense to us?

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    $\begingroup$ The way we happen to plot real and complex numbers is due to convenience. If you want to plot real numbers on a wiggly line go ahead. For instance, there's a 1-1 correspondence between the points on a sine curve and the real numbers. So it shouldn't make any difference if you, for some reason, decided to plot the real numbers there. $\endgroup$
    – user137731
    Commented Jun 28, 2016 at 20:56
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    $\begingroup$ If you interpret multiplication by a complex number geometrically, multiplication of a complex number $v$ by $i$ is equivalent to rotating $v$ by $90^\circ$ counter-clockwise. Before even knowing that it should be $90^\circ$, we know that the rotation should be the same going from $1$ to $1\cdot i$ as it should be from $i$ to $i\cdot i$. Since $i^2=-1$, it makes sense that it should be $90^\circ$ rotation. $\endgroup$
    – JMoravitz
    Commented Jun 28, 2016 at 20:57
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    $\begingroup$ You can bend and twist the complex plane (or already the real line) as much as you like. However, in that case the arithmetic operations of addition and multiplication do not correspond to simple geometric transformations any more. If you can live without that, fine. After all, the real line/the complex plane is only a model for and not identical to the field of real/complex numbers $\endgroup$ Commented Jun 28, 2016 at 20:58
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    $\begingroup$ I wouldn't say "arbitrary". I'd prefer "convenient". $\endgroup$
    – user137731
    Commented Jun 28, 2016 at 21:01
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    $\begingroup$ John Stillwell, in Mathematics and Its History, 1st ed., pp. 191-192, credits the first attempt at a geometric representation of complex numbers to Wallis, and points out that it (1) was very different from the usual modern graphing of complex numbers and (2) didn't work well. The conventional mapping of the complex numbers onto a Euclidean plane is just so incredibly helpful (with geometric analogies to complex operations) that it's easy to forget that it is not part of the definition of complex numbers at all. $\endgroup$
    – David K
    Commented Jun 29, 2016 at 3:35

11 Answers 11


There is a really important aspect of complex numbers that depends on the complex plane having exactly this shape: complex multiplication.

Complex numbers can not only be characterized in cartesian coordinates by a real part and an imaginary part, but also in polar coordinates by a length and an angle.

You know that for any $z \in \mathbb{C}$ there exist $x, y \in \mathbb{R}$ such that $z = x+i\cdot y$, right? $x$ is the real part and $y$ is the imaginary part? Well, there also exist $r, \varphi \in \mathbb{R}$, $r \geq 0$ such that $z = r\cdot(\cos\varphi + i\sin\varphi)$. Here, $r$ is called the length or absolute value of $z$ and $\varphi$ is called the angle or argument, measured counterclockwise from the positive real axis.

We can use cartesian coordinates to add complex numbers: $$(x_1+i\cdot y_1) + (x_2+i\cdot y_2) = (x_1+x_2) + i\cdot(y_1+y_2)$$

We can use cartesian coordinates to multiply complex numbers:

$$(x_1+i\cdot y_1)\cdot (x_2+i\cdot y_2) = (x_1x_2-y_1y_2) + i\cdot(x_1y_2+y_1x_2)$$

However, we can also use polar coordinates to multiply complex numbers:

$$(r_1(\cos\varphi_1 + i\sin\varphi_1))\cdot(r_2(\cos\varphi_2 + i\sin\varphi_2)) = (r_1\cdot r_2)(\cos(\varphi_1+\varphi_2) + i\sin(\varphi_1+\varphi_2))$$

So to multiply two complex numbers in polar coordinates, you multiply their lengths and add their angles. I personally think this is incredibly helpful for visualization, and this also shows why the imaginary axis needs to be at a right angle to the real axis: since the angle of $-1$ is $180^\circ$, the angle of $i$ needs to be $90^\circ$ or $270^\circ$.

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    $\begingroup$ Worth mentioning that this also leads to Euler's formula and simplifies the way that complex exponentiation works. $\endgroup$
    – fluffy
    Commented Jun 29, 2016 at 1:43

The complex plane is a model (or more formally, an isomorphism) with very nice properties, such as $\arg(zw)=\arg(z)+\arg(w)$ or, in more advanced complex analysis, winding numbers.

Could there be a "better" representation? Maybe. And, in any case, you should be clear about what means "better" for you, that is, what you want to do with this representation.

What is sure is that, since Gauss began to use the complex plane, deep, useful and beautiful applications have been found.

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    $\begingroup$ This answer is so insanely good. Seriously. To anyone reading this: take a moment and let it all sink in. $\endgroup$
    – The Count
    Commented Jun 28, 2016 at 21:11
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    $\begingroup$ @TheGreatDuck Typically, it is used to denote the angle from the positive real axis of your complex number. For example, $\arg(1+i)=\frac{\pi}{4}$. $\endgroup$
    – The Count
    Commented Jun 28, 2016 at 21:14
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    $\begingroup$ To play devil's advocate, it seems that without having already decided on the plane, the arg function looks rather arbitrary. Moreover, winding numbers are purely topological, so they don't tell us anything about the geometry of the plane. (Of course, if we accept that $e^x$ is a nice function, then things like the argument pop out, but that's not mentioned in this answer) $\endgroup$ Commented Jun 28, 2016 at 22:29
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    $\begingroup$ and the last sentence is rather euphuistic $\endgroup$
    – miracle173
    Commented Jun 28, 2016 at 22:33
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    $\begingroup$ This is really not that useful... $\endgroup$
    – Pedro
    Commented Jun 29, 2016 at 17:16

We choose pictorial representations so that algebraic properties (such as associativity and commutativity of addition and multiplication) have coherent geometric interpretations.

For example, the traditional number line associates to each real number $a$ a displacement from $0$ in such a way that for all real $a$ and $b$,

  • Addition corresponds to concatenation of displacements, subtraction $b - a$ is the displacement taking $a$ to $b$, and $|b - a|$ is the distance from $a$ to $b$;

  • Multiplication by a positive real $c$ corresponds to scaling by $c$ (fixing $0$) and multiplication by $-1$ is the (distance-preserving) reflection across $0$;

  • Moving rightward corresponds to the order relation.

There are, however, other geometric ways to represent the field of real numbers. To name one, identify the real number $a$ with the line of slope $\tanh a$ through the origin in the Cartesian plane, and let the reals act by boost transformations. This is how velocities add in special relativity.

The Euclidean plane [*] can be coordinatized by fixing an origin $O$, two mutually-perpendicular oriented lines $X$ and $Y$ through $O$, and a unit of length. (Conventionally, we take $X$ horizontal and oriented to the right; $Y$ vertical and oriented upward.) If $P$ is an arbitrary point, there is a unique line $\ell_{Y}$ through $P$ and parallel to $Y$ by the parallel postulate; the unique point of intersection of $\ell_{Y}$ with $X$ defines the $x$-coordinate of $P$. The $y$ coordinate of $P$ is defined similarly.

The point $P$ is thereby identified with an algebraic address: the ordered pair $(x, y)$ of real numbers. Doing so furnishes a bijective correspondence between the Cartesian plane (the Euclidean plane with the extra structure of the origin, two oriented lines, and a fixed unit of length) and the set $\mathbf{R}^{2}$ of ordered pairs of real numbers.

Because each axis is effectively a number line, the algebraic operations of vector addition acquire geometric meaning, via the parallelogram law for vector addition and the operation of radial scaling centered at the origin for scalar multiplication.

Now, what about complex numbers? Identifying $\mathbf{C} \simeq \mathbf{R}^{2}$ is natural enough: $x + iy \leftrightarrow (x, y)$. Because $i^{2} = -1$, we wish to represent $i$ as a geometric operation that, performed twice in succession, has the effect of multiplying by $-1$, i.e., of rotating the plane by a half-turn about the origin. A quarter turn about the origin accomplishes this. By an additional convention, we (usually) take the counterclockwise rotation, so that the positive $x$-axis rotates to the positive $y$-axis. (We could just as consistently have taken the clockwise rotation; the algebraic operation of complex conjugation is a field automorphism.)

Admittedly this is a compressed account; the essential ideas, however, may be easier to see if not every detail is filled in. The bottom line is, the correspondence between algebra and geometry is conventional, though (as other answers note) mathematically rich beyond any reasonable a priori expectation.

[*] In the original post, I wrote, "Descartes noticed that the Euclidean plane...." As Daniel R. Collins noted in the comments, this assertion is less true than "common knowledge" has it. I'm not a mathematical historian, and can't lay out the complete provenance of Cartesian axes, but axes as described in this answer were not explicitly introduced by Descartes. (The comments contain a few more details.)

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    $\begingroup$ My understanding is that Descrates made use of a one-dimensional line, but not two perpendicular axes; that is properly credited to a later translator of his work, Frans van Schooten, et. al. Suggest either supplying a citation for that use by Descartes or removing his name from that paragraph. $\endgroup$ Commented Jun 29, 2016 at 1:40
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    $\begingroup$ @DanielR.Collins: I haven't read Descartes in French, but the assertion that Descartes didn't coordinatize the plane is new to me. In A History of mathematics, for example, Florian Cajori (p. 215, ten lines from the bottom) writes, "By him [Descartes], a point on a plane was determined in position by its distance from two fixed right lines or axes." If this well-trodden claim is incorrect, I'd certainly like to know. Do you have sources to the contrary? $\endgroup$ Commented Jun 29, 2016 at 1:57
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    $\begingroup$ First place I found it was on Wikipedia, with a citation of Burton, David M. (2011), The History of Mathematics/An Introduction (7th ed.), p. 374 (en.wikipedia.org/wiki/Cartesian_coordinate_system). Second place is in printed lecture notes by George Smith, Tufts U., Philosophy 167: Science Before Newton’s Principia, Class 8, on Descartes' Geometry: "Though not a reduction of geometry to algebra (just the opposite) and no Cartesian coordinates in it" $\endgroup$ Commented Jun 29, 2016 at 4:23
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    $\begingroup$ @DanielR.Collins: Interesting, thanks. The only source to which I have immediate access is the 1886 A. Hermann edition of La Géométrie. At a quick skim I don't see explicit introduction of "axes", but (for example) on page 33 (Fig. 12, ff.), Descartes (modulo my non-fluent French) describes a hyperbola using displacements $x$ and $y$ along mutually perpendicular lines. By chance, the Clavius Group is meeting this week; I'll ask around.... $\endgroup$ Commented Jun 29, 2016 at 11:49
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    $\begingroup$ @DanielR.Collins: Thank you for the correction. :) I've removed the phrase "Descartes noticed that..." and added an explanatory footnote. $\endgroup$ Commented Jun 30, 2016 at 21:34



made it into a sphere:


  • $\begingroup$ You. Just. Blew. My. Mind. Thank you for the great bit of information! $\endgroup$
    – user64742
    Commented Jun 30, 2016 at 17:20
  • $\begingroup$ Don't worry. They never taught us what the point of this is. They merely stated its existence and then went on with other stuff. $\endgroup$
    – null
    Commented Jun 30, 2016 at 17:28
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    $\begingroup$ I am not worried. $\endgroup$
    – user64742
    Commented Jun 30, 2016 at 17:51
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    $\begingroup$ @Bye_World: Well, among other things, objects that are normally lines become circles through the point at infinity. Which simplifies a lot of theorems so instead of saying "lines do this and circles do that..." you can just say "circles do that..." and it covers both cases. $\endgroup$ Commented Jul 1, 2016 at 3:17
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    $\begingroup$ @TheGreatDuck, should you have time, look up "stereographic projection". $\endgroup$ Commented Jul 1, 2016 at 3:21

This is a long-winded version of some of the answers already posted. Hopefully it helps someone. Be sure to take a look at the comments below for slightly more sophisticated approach.

Forget everything you know about the complex number system $\mathbb{C}$ for a moment. Let's turn the question on it's head: suppose we have a two dimensional number system $\mathbb{K} = \{a+b\kappa:a,b\in\mathbb{R}\},$ and suppose our new number $\kappa\not\in\mathbb{R}$ is "perpendicular to the real axis," in a sense that we shall soon define precisely. Suppose also that multiplying these numbers behaves nicely, in a precise way. In this answer, we shall show that $\kappa^{2}=-1$ then must be the case. The point is that the complex numbers $\mathbb{C}$ are not just introduced for no reason (or for some boring reason like quadratic equations): it is useful because it is the only number system which behaves nicely enough for our usual purposes. The answer to the question is then "the complex plane is shaped like it is because we want complex numbers to be elegant".

First, by "a number system", I just mean a set of objects which we can add to each other, subtract from each other, multiply by each other and divide by each other (except by zero!), just like with real numbers.

Next, what do I mean by "perpendicular to the real axis"? We want to be able to draw $\mathbb{K}$ as a plane, with the real axis horizontal and the $\kappa$ axis vertical. Imagine you've drawn the plane (or better yet, actually draw the plane) and pick a point $a+b\kappa\in\mathbb{K}$. This point lies somewhere in the plane. Taking inspiration from our knowledge of the Cartesian plane $\mathbb{R}^{2},$ we define notions of length and angle (or modulus and argument) corresponding to $a+b\kappa$ as follows:

  • Since the point $a+b\kappa$ naturally corresponds to the point $(a,b)\in\mathbb{R}^{2}$, and the length we typically associate with $(a,b)$ is the Euclidean distance to the origin, we define the modulus of $a+b\kappa$ by $$\lvert a+b\kappa \rvert = \sqrt{a^{2}+b^{2}}.$$
  • Since the point $a+b\kappa$ naturally corresponds to the point $(a,b)\in\mathbb{R}^{2}$, and the angle we typically associate with $(a,b)$ is the angle to the real axis measured counterclockwise, we define the argument of $a+b\kappa$ by $$\arg{(a+b\kappa)} =\begin{cases}\tan^{-1}{(b/a)} & \text{if }a\neq0\text{ and }b>0,\\-\tan^{-1}{(b/a)} & \text{if }a\neq0\text{ and }b<0,\\\pi/2 & \text{if }a=0\text{ and }b>0,\\-\pi/2 & \text{if }a=0\text{ and }b<0\\0&\text{if }a>0\text{ and }b=0\\\pi&\text{if }a<0\text{ and }b=0.\end{cases}$$ Note that if $a=b=0$, then we leave the argument undefined.

At this point you should probably draw a sketch and make sure this agrees with what we know about coordinate geometry. This definition guarantees in advance that $\arg{\kappa}=\pi/2$, which is the whole point. Also, $\lvert\kappa\rvert=1$ by definition.

Finally, what do I mean by "multiplication of these numbers behaves nicely"? I mean this: multiplication of two numbers in $\mathbb{K}$ should result in a number whose length is the product of their lengths and whose argument is the sum of their arguments.

I understand that this last assertion might seem a bit random, so I'll try to convince you that this is a reasonable thing to impose. Note that multiplying any number $a+b\kappa\in\mathbb{K}$ by a positive real number $r>0$ a priori does not affect the argument: $$\arg{(r(a+b\kappa))}=\tan^{-1}{\frac{rb}{ra}} = \tan^{-1}{(b/a)} = \arg{(a+b\kappa)}.$$ This means that multiplying $a+b\kappa$ by a positive real number $r$ gives a number whose argument is $\arg{(a+b\kappa)}+0$, and we note here that $0=\arg{r}$. Similarly, multiplying by negative real numbers results in addition of $\pi$ to the argument, which is handy because negative real numbers have argument $\pi$. When it comes to multiplying two arbitrary non-zero numbers in $\mathbb{K}$ together, we would like there to be a rule that fits the current pattern. I've already hinted at the most obvious one: $$\text{if }z,w\in\mathbb{K}\setminus\{0\},\text{ then }\arg{(zw)}=\arg{z}+\arg{w}.$$ By very similar reasoning, it makes sense to impose the rule $\lvert zw\rvert =\lvert z\rvert \lvert w\rvert$.

Side note: hopefully you agree that these are nice properties for our number system to have. The whole point of this discussion is that it would be aesthetically pleasing (and, as it turns out, useful!) to have a number system that follows these rules.

Now, if $\mathbb{K}$ truly behaves as we want it to and follows all the rules we've laid out above, then we have no choice: since $\lvert\kappa\rvert=1$ it follows that $\lvert\kappa^{2}\rvert=1$ and $\arg{\kappa^{2}}=\pi/2+\pi/2=\pi$, and the only number with length $1$ and argument $\pi$ is $-1$. Hence $\kappa^{2}=-1$, and $\mathbb{K}=\mathbb{C}$.

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    $\begingroup$ There is perhaps another, slicker, way of looking at this. If we impose only that arguments add together, and we say $\kappa^{2}=e+f\kappa$ for some $e,f\in\Bbb{R}$, then, if $a\neq0\neq c$, calculating the argument of $(a+b\kappa)(c+d\kappa)$ leads us to conclude, by the arctan summation formula, that $$\frac{ad+bc}{ac-bd}=\frac{ad+bc+bdf}{ac+bde}.$$ Now setting $e=-1,f=0$ gives the simplest solution. Unfortunately, I couldn't see a way to show that this is the only solution (and I'm not convinced it is), so I abandoned this approach. $\endgroup$
    – Will R
    Commented Jun 29, 2016 at 1:44
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    $\begingroup$ If that equation is true for all $(a,b,c,d)$ (or even for some nonempty open subset of $\Bbb R^4$), then it is certainly the case that $e=-1,f=0$. If you cross multiply you get $(b^2cd+abd^2)(e+1)+(b^2d^2-abcd)f=0$, and the surfaces $b^2cd+abd^2=0$ and $b^2d^2-abcd=0$ are nowhere dense (and neither is a subset of the other), so you can pick points on each surface and not the other to show $e+1=0$ and $f=0$. $\endgroup$ Commented Jun 30, 2016 at 3:59
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    $\begingroup$ @MarioCarneiro: Nice! So that means that if we want to define a "natural" multiplication on $\mathbb{R}^{2}$ (using the standard basis and "multiplying out the brackets") for which $\arg$ has this kind of completely additive property, then we have no choice: we must choose $(0,1)^{2}=(-1,0)$, giving $\mathbb{C}$. In some sense this provides motivation for the standard construction of $\mathbb{C}$; or at least, it's a somewhat persuasive algebraic reason for $\mathbb{C}$ to be worthy of our attention from the get-go. $\endgroup$
    – Will R
    Commented Jun 30, 2016 at 12:20

Complex numbers can be constructed as couple of real numbers : $ a+ib=(a,b) $ with suitable definitions of the operations of sum and product ( see here). With such definitions a complex number corresponds, in a natural way, to an element of $\mathbb{R}^2$ and we have : $1=(1,0)$ and $i=(0,1)$ and , using the usual representation of an orthogonal system for the coordinates in $\mathbb{R}^2$, these two vectors, are represented as points, at distance $=1$ from the origin, on two straight lines that forms a $90°$ angle. So this same representation is also adopted for the complex numbers.


Because of complex conjugation, the complex numbers have a natural measure of distance; the "size" (i.e. distance from zero) of a complex number is

$$ \| z \| = \sqrt{z \bar{z}} $$

The right hand side is well-defined in terms of more basic concepts, because $z \bar{z}$ is a nonnegative real number.

More generally, the distance between any two complex numbers is

$$ \text{Distance}(z, w) = \|z - w\| $$

which is enough to reproduce plane geometry. e.g. we can see that $1$ and $i$ are at right angles, because they describe two of the sides of a right triangle with $0$ as the remaining vertex:

$$ \|1\|^2 + \|i\|^2 = \|1-i\|^2 $$

Furthermore, the two main ways of expressing complex numbers can be motivated for purely algebraic reasons. It is convenient to combine conjugation with addition to split a number into two parts

$$ 2z = (z + \bar{z}) + (z - \bar{z}) $$

The first and second terms ultimately lead to the usual real and imaginary parts of $z$.

Combining with multiplication instead suggests, for $z \neq 0$,

$$ z^2 = (z \bar{z}) \cdot \frac{z}{\bar{z}} $$

The first term leads to the norm as described earlier. The second term leads to reducing $z$ to a "unit vector" $\frac{z}{\|z\|}$ on the circle, and thus basically corresponds to the complex argument (i.e. angular position relative to the origin).


1. $90$ degrees is chosen because it geometrically matches what was already working. -1 is 180 degrees around from $1$. When you multiply by $-1$ you rotate $180$.
and $$(-1) \cdot( -1)=1$$
It works the same way for $i$ except $90$ degrees.
$$1\cdot i=i$$ and
$$i\cdot i=-1$$ and

2. Has to do with radians. Radians are more mathematical versions of degrees.
$$360\; \mathrm{degrees}=2\pi\;\mathrm{radians}$$ and

$$re^{\theta i}=r(\cos(\theta)+i\sin(\theta))$$ In this way of showing imaginary numbers $i=e^{\frac{\pi}{2}i}$, so this means that $i$ is literally $90$ degrees from $1$ and $-1$.


Sometimes a relationship between something algebraic and something geometric is almost magical. A simple example would be the linear functions y = mx + b and graphing these functions to obtain lines in a cartesian plane. Just about any interesting algebraic observation you might make about y=mx + b has a clear geometric counterpart about lines.

The realization of the complex plane as, well, a complex plane is such a relationship. Sure you could put the imaginary axis at a non-90 degree angle, but then the relationships between the geometry of your plane and the algebra of complex numbers would all become much less transparent.

Take the group of numbers cos(a) + i sin(a). These form an algebraic group, and with the usual realization they are a circle, and multiplying by one of them is a rotation of the complex plane. However, if you change the angle of the imaginary axis, then your circle is replaced by an ellipse at some skew angle from the real numbers, and multiplying by these numbers does odd deformations of the complex plane instead of simply rotating it.

You can think of the sun as rotating around the earth if you want to, but then physics looks very confusing. When you find how much simpler physics looks when you see the earth as rotating around the sun, you conclude that must be the way things really work.

  • $\begingroup$ Eh but that's only from the viewpoint of the sun. From the viewpoint of the galactic centre the sun and earth are going around it, and their orbits weave in and out of each other, not at all looking like the earth going round the sun. =) $\endgroup$
    – user21820
    Commented Jul 2, 2016 at 4:49

Why are complex numbers graphed like points in $\mathbb{R^2}$? Because we choose too, and as a result of our choices (through trigonometry and algebra) the complex numbers became a very useful and beautiful group of numbers.

We defined the complex numbers. Both algebraically and geometrically. And choose to define the pure complex unit by:


Through the Taylor series and the definition of $i$ we got this astonishing formula.

$$re^{i\theta}=r\cos (\theta)+ir\sin (\theta)$$

In a coordinate system with two perpendicular axis $\mathbb{R^2}$, $r\cos(\theta)$ and $r\sin(\theta)$ are given meaning as we recognize these expressions as components of a triangle with hypotenuse $r$ and angle $\theta$ from the origin. The cosine expression being the horizontal component and the sine being the vertical one. But only in such coordinate system will our trigonometry and geometric knowledge make simple sense. So we decided to use this coordinate system $\mathbb{R^2}$ and as a result of the convenience of $r\cos (\theta)$ and $r\sin (\theta)$, we choose to graph the real component of our complex number in the horizontal axis and the imaginary components in the vertical axis.

And more surprises came when we looked at the geometry of multiplying complex numbers. We graph complex numbers as a whole, thus multiplying by $1$ will get you back to the same complex number, so it's the equivalent of a $360$ degree turn. But multiplying by $-1$ will only get you back to the same complex number if you do it twice, so it's the equivalent of half a $360$ turn, I.e a $180$ degree turn. Multiplying by $i$ four times gets you back to the same number so it's the equivalent of a $90$ degree turn....and furthermore every complex number is given a rotational sense when multiplication is performed.

So as a result of our choices, our choices of how we defined the complex numbers and how we graphed them, we open a door to a brand new set of mathematical tools.

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    $\begingroup$ You say "we choose to define the pure complex unit by: $i^2=−1$. And choose to keep all the same properties (associative, communitive, distributive) over addition and multiplication. And choose to have multiplicative inverses for all complex numbers except $0$, $0$ remained the additive identity, and $1$ remained the multiplicative identity.". This is wrong. You cannot any how choose what you want. You've to first prove that such an object exists. This kind of invalid reasoning is the core of almost all known paradoxes. If what I say isn't clear to you, I suggest you learn logic. All the best. $\endgroup$
    – user21820
    Commented Aug 1, 2016 at 11:42

for the beginning we will work on $\mathbb{T}=\{z \in \mathbb{C} ; |z|=1\}$ only because we need to explain the geometric meaning of argument only.

so we begin with what we know all (real numbers) and solving the equation of type $ax^2+bx+c=0$, we know all using canonical form that this equation can be transformed to $y^2=t$ where $t$ is a real number, so we need to explain what is the effect of the square function to a number.

so we draw the real line and we see how this function transform a real number of $\mathbb{T}$ ($1$ and $-1$) first for $1$ we see that the angle which form the vector $1$ (think about the vector whose extremities are $0$ and $1$) is $0$, the application $x\to x^2$ transform $1$ to 1 so the angle stay $0$, but for $-1$ we see that the angle which form the vector $-1$ is $\pi$ and $(-1)^2=1$ so the angle is $2\pi\equiv 0$, so as conclusion we can say that the function square "redoubles the angle" of the vector induced by a number.

as a direct consequence we can say that the equation $x^2=-1$ can't have a real solution because the angle of $x^2$ will be $\pi$ so by the above result the angle of $x$ will be $\pi/2$ which is not a real number, so call this number $i$ and we can define the complex plane as the linear combination of this two "independent" vectors.


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