Why imaginary numbers axis is plotted perpendicular to the real numbers axis?

Question

Negative numbers axis is plotted to the opposite side of the positive real number axis that make sense but i do not understand why imaginary numbers are plotted perpendicular to the real numbers axis.

If you do it that way, then addition and multiplication have nice geometric interpretations. For example, to multiply two complex numbers, you add the angles and multiply the lengths. (As a special case, multiplying by $ i $ simply rotates by 90 degrees counter clockwise. ) — littleO, May 26 '15 at 08:39
Part of that is De Moivre, I guess. The theorem that:$$(\cos A+i\sin A)(\cos B+i\sin B)=\cos(A+B)+i\sin(A+B)$$has the geometric interpretation of adding angles. ($\cos A+i\sin A$ is the point on the unit circle at angle $A$.) Also, the norm, $\lvert x+iy\rvert=\sqrt{x^2+y^2}$ has the geometric interpretation of distance from the origin, so the theorem that $\lvert w\rvert\lvert z\rvert=\lvert wz\rvert$ has the interpretation of multiplying distances. — Akiva Weinberger, May 26 '15 at 09:05
Are you trying to bring into public debate the political correctness of using a straight angle as the foundation stone for multidimensional extensions of the real numbers ? :-$)$ — Lucian, May 26 '15 at 11:45
Sorry dudes the best answer that i have got is that just as -1 itself can be thought of as 1 at an angle of 180 degrees so The square root of -1 just can be thought of as 1 at an angle of 90 degrees. Is not that one is simplest ? @littleO — hur chu chu, May 27 '15 at 12:18
First 1 and i are linearly independent over R from which C is isomorphic to R ^ 2 and C can be represented as R ^ 2 of the way you want. But more suitable is the perpendicularity of the axes. — Piquito, May 27 '15 at 13:18

score 4 · Answer 1 · answered May 26 '15 at 08:42

4

The most elementary complex number $ i$ equals $ e^{i \pi/2} $ by Euler's theorem. So it is natural to take $ \theta = \pi/2 $ line for imaginary number axis on a line perpendicular to real axis where real component=0 or origin.

answered May 26 '15 at 08:42

Narasimham

40,495

Mr. Proof · Answer 2 · 2021-08-03T14:08:04.030

3

Vey nice question. Think of the solution of the equation $x^2+1=0$, there is not real solution in the real line (real-world), but the polynomial has to have a root! so the solution must be in a different world, so mathematicians extended the real world to infinitely many real worlds $x$ shifted by some non real world $yi$. So we got $x+yi$ where our actual world (real line) just a single slice of the extended world (real + imaginary).

Now, why do we plot it perpendicular? If we do not do that means for every real number can be represented by an imaginary numbers (or vise versa) by using Pythagorean relation which means that the imaginary and real numbers are just the same which is against the task of extending the real line to complex plane.

edited Aug 03 '21 at 14:08

answered Jul 29 '21 at 01:40

Mr. Proof

1,477

2

Its not an intuitive explanation. Its like calculation value of pi from a lot of different equations. I want to know how did the early mathematicians connected complex numbers with geometry. De-Morgan's law is only proved by induction, it doesn't give the understanding or clear picture of equation. – Usercomingsoon Jul 29 '21 at 11:37
2

@Usercomingsoon If you want some relevant history read (at least) the one paragraph in the article Stanford Encyclopedia of Philosophy: René Descartes where the word 'coordinate' appears. – CopyPasteIt Jul 31 '21 at 09:01
1

@Usercomingsoon The huge leap was the Euler identity, namely $e^{i x} = \cos x + i \sin x$. It is the connection between torus $\mathbb{T} = { z \in \mathbb{C}; |z| \le 1}$ and real-frequency $sin x$ and imaginary- frequency $i \cos x$. In fact it is not intuitive. – Mr. Proof Aug 01 '21 at 12:46
1

The 0! might make someone think of the factorial where 0!=1. Other than that, I like this answer. – Тyma Gaidash Aug 02 '21 at 12:35
@TymaGaidash you are right. – Mr. Proof Aug 03 '21 at 14:07

5xum · Answer 3 · 2015-05-26T08:38:49.980

2

One (not the only) good reason: if you define the norm of $z\in\mathbb C$ as $$||z|| = \sqrt{z\bar z}$$

then $\mathbb C$, as a normed vector space over $\mathbb R$, is isomorphic to $\mathbb R^2$.

edited May 26 '15 at 08:38

answered May 26 '15 at 08:34

5xum

123,496
6
128
204

score 1 · Answer 4 · answered May 27 '15 at 12:16

1

Sorry dudes the best answer that i have got is that just as -1 itself can be thought of as 1 at an angle of 180 degrees so The square root of -1 just can be thought of as 1 at an angle of 90 degrees. Is not that one is simplest ?

answered May 27 '15 at 12:16

hur chu chu

191

2

Yeah, pretty much. But that's a special case of a more general theorem: "$1$ at an angle of $\alpha^\circ$" times "$1$ at an angle of $\beta^\circ$" is equal to "$1$ at an angle of $\alpha+\beta^\circ$". (Which can be used to derive certain identities in trigonometry.) – Akiva Weinberger May 27 '15 at 12:36
1

Despite of $i=\sqrt{-1}$, I guess the real number -1, as well as a rotated 1, has very few to tell in this story. – Piquito May 27 '15 at 15:08

score 0 · Answer 5 · answered Jul 29 '21 at 01:23

In modern mathematics, the abstract Cartesian Coordinate Plane is a an underlying set

$\quad \Bbb R \times \Bbb R$

with additional properties that can 'flesh it out' and give it more structure.

In set theory there is no such thing as plotting points, but every math student has graphed functions and relations on graph paper using a pencil; every point on the paper corresponds to an element of the set $\Bbb R \times \Bbb R$.

Without a doubt, the student knows that $x\text{-axis}$ numbers increase as you move right and the $y\text{-axis}$ numbers increase as you move up; that is just the convention. But in this wiki article you'll read

(However, in some computer graphics contexts, the ordinate axis may be oriented downwards.)

When if comes to the complex numbers, the student learns that,

$\quad \Bbb C = \{x + yi \mid \text{where } x \in \Bbb R \text{ and } y \in \Bbb R \}$

and

$\quad a + bi = c + di \text{ iff } a = c \land b = d$

But then the complex numbers can be plotted on graph paper - plot $x + yi$ to correspond to the element $(x,y) \in \Bbb R \times \Bbb R$.

We could also use a different bijective correspondence,

$\quad x +iy \mapsto (x,-y) \quad$

but we agree, by convention, to have $i$ point orthogonally up from the $x\text{-axis}$
(note also that both $i$ and $-i$ satisfy $x^2 = -1).$

As the student learns how to graph the binary operations of addition and multiplication of complex numbers, in no time at all they will see that the identification of $\Bbb C$ with $2\text{-dimensional space}$ has its merits.

score 0 · Answer 6 · answered Aug 02 '21 at 03:25

The Complex plane is technically $\mathbb{R}^{2}$ equipped with a new operation and elements. Such is why $\mathbb{C}$ is plotted on the familiar $y$-axis.

One can construct $\mathbb{C}$ as follows:

Let vectors $\vec{e}_{1}=(1, 0)$ and $\vec{e}_{2}=(0, 1)$. Write every vector $\vec{z}\in \mathbb{R}^{2}$ as $\vec{z} = (x, y) = x\vec{e}_{1} + y\vec{e}_{2}$, such that $\vec{e}_{1}$ and $\vec{e}_{2}$ form a new basis.

Finally, define the multiplication operation on our constructed plane with $\vec{e}_{1}^{2} = \vec{e}_{1}$, $\vec{e}_{2}^{2} = -\vec{e}_{1}$, $\vec{e}_{1}\vec{e}_{2} = \vec{e}_{2}\vec{e}_{1} = \vec{e}_{2}$. We denote $\vec{e}_{1}$ by $1$, and $\vec{e}_{2}$ by $i$.

The standard complex number addition and multiplication operations follow from these definitions.

Why imaginary numbers axis is plotted perpendicular to the real numbers axis?

6 Answers6