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I was just wondering how did people know that the distance between $0$ and $i$ is $1$ in the complex plane, did they just assume this, is it just an axiom, or is there a proof behind it or a reason for it?

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    $\begingroup$ I suppose it's just a convention to put $i$ at a distance $1$ from the origin in the complex plane; but it's nice to have $|zw|=|z||w|$ where $|z|$ denotes the distance from $z$ to the origin, don't you think? $\endgroup$ Nov 3 '18 at 20:34
  • $\begingroup$ In some sense, I think the ordinate of a point $z$ at complex plane is exactly its imaginary part, by definition $b\in\mathbb{R}$ such that $z=a+bi$ ($a\in\mathbb{R}$ as well). I understand the answers below, by in some sense I thinks it's a definition (maybe some answer are ciclic). I'd like to receive answer to this comment... I'm curious now. Thanks. $\endgroup$ Nov 3 '18 at 20:35
  • $\begingroup$ How did people know that the distance between $0$ and $1$ is $1$ in the complex plane? Same thing. $\endgroup$
    – MPW
    Nov 3 '18 at 20:45
  • $\begingroup$ How did we know that the distance from $(0,1)$ to $(0,0)$ = $1$? $\endgroup$
    – John Douma
    Nov 3 '18 at 20:55
  • $\begingroup$ @LordSharktheUnknown I think that is the heart of it. As a real vector space the complex numbers can be regarded as an affine system, and one can create a metric using any pair of basis vectors as units. But if you want the metric to be compatible with multiplication of complex numbers then there is only one choice. $\endgroup$ Nov 3 '18 at 21:55
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We define the distance between $x$ and $y$ in the complex plane as being $||x-y||$ where if $||a+bi||^2 = a^2+b^2$. Substitution gives that the distance between $i$ and $0$ is $||i-0||=||i||=1^2=1$. However, this doesn't really explain why this is true. However, if you are familiar with vectors the formula $||x-y||$ should look familiar. In fact, this is just the same distance formula we use in the plane. So what happened is that we decided that the distance between $a+bi$ and $c+di$ should be the same as the distance between $(a,b)$ and $(c,d)$. In some sense, this is the more fundamental assumption about what we mean by distance in the complex plane, and then the fact that the distance from $0$ to $i$ is $1$ follows from that.

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    $\begingroup$ The first four words cannot be overemphasized $\endgroup$ Nov 3 '18 at 20:34
  • $\begingroup$ so there isn't really a proof? just an axiom? $\endgroup$ Nov 4 '18 at 19:56
  • $\begingroup$ @MohannadEl-Nesr. There is no proof of the definition, but given the definition, there is the proof outlined above (ie. the distance between $i$ and $0$ is $||i||=1^2=1$) $\endgroup$
    – memerson
    Nov 4 '18 at 20:59
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Note that $i=(0,1)$ and the origin is $(0,0)$

According to the distance formula we get $d= \sqrt {0^2+1^2} =1$

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  • $\begingroup$ +1 What are we missing? This seems trivial. We knew because we defined the complex numbers. $\endgroup$
    – John Douma
    Nov 3 '18 at 20:58
  • $\begingroup$ @JohnDouma exactly, it's "defined" but I am searching for a proof, obviously it's just convention from the answers I've been seeing but really stating the distance formula doesn't help. $\endgroup$ Nov 20 '18 at 12:38
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This is, as mentioned by others, a matter of definition, but there is a reason to believe that the definition is correct.

Given a real polynomial $p(x)=(x-a_1)\cdots(x-a_n)$ with the $a_i$ real, then, when $a$ is a real number such that $p(a)\neq 0,$ the radius of convergence of the Taylor series of $1/p(x)$ at $x=a$ is $\min_i |a-a_i|.$

We can also show[*] that the Taylor series for the real function $f(x)=\frac1{1+x^2}$ at $x=a$ has interval of convergence with radius equal to $\sqrt{1+a^2}.$

This gives the impression that, if there there are roots of $x^2+1,$ they must be a distance $1$ from $0$ in a direction perpendicular to the real line.

So, even before we have the complex numbers, we know roughly “where” we’d want the root(s) to $x^2+1$ to exist, if they exist.


[*] It is a tedious argument if kept in the reals, but it can be done.

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The distance between two complex numbers $z$ and $z'$ is the module of the difference $|z-z'|$. In particular, the distance to $0$ is the module.

Now $|z|^2=z\bar z$, so $|i|^2=i(-i)=1$, hence $|i|=\sqrt 1$.

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We simply have by definition

$$|i|=\sqrt{1^2}=1$$

enter image description here

(credit Wikipedia)

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$\mathbb{C}=\{x+iy:x,y\in\mathbb{R}\}$ where $i$ is a root of the equation $z^2+1=0$. A metric $d:\mathbb{C}\times\mathbb{C}\rightarrow \mathbb{R}_{\geq 0}$ is defined by $d((x_1+iy_1),(x_2+iy_2))=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ for $x_j,y_j\in\mathbb{R},j=1,2.$ This metric is called the natural distance in $\mathbb{C}$ because of its similarity with the distance in $\mathbb{R}^2.$ Now by this definition, distance between $0$ and $i$ is $d((0+0i),(0+1i))=\sqrt{0+1}=1.$

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  • $\begingroup$ Thanks! That was a typo. Edited now. $\endgroup$
    – Surajit
    Nov 4 '18 at 18:57
  • $\begingroup$ You also have $x_2$ and $y_1$ swapped. $\endgroup$
    – mr_e_man
    Nov 5 '18 at 17:11

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