# Is there a proof that the distance from $0$ to $i$ in the complex plane is $1$?

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?

• 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? Nov 3 '18 at 20:34
• 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. Nov 3 '18 at 20:35
• How did people know that the distance between $0$ and $1$ is $1$ in the complex plane? Same thing.
– MPW
Nov 3 '18 at 20:45
• How did we know that the distance from $(0,1)$ to $(0,0)$ = $1$? Nov 3 '18 at 20:55
• @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. Nov 3 '18 at 21:55

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.

• The first four words cannot be overemphasized Nov 3 '18 at 20:34
• so there isn't really a proof? just an axiom? Nov 4 '18 at 19:56
• @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 $||i||=1^2=1$) Nov 4 '18 at 20:59

Note that $$i=(0,1)$$ and the origin is $$(0,0)$$

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

• +1 What are we missing? This seems trivial. We knew because we defined the complex numbers. Nov 3 '18 at 20:58
• @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. Nov 20 '18 at 12:38

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.

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 $$|i|=\sqrt 1$$.

We simply have by definition

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

(credit Wikipedia)

$$\mathbb{C}=\{x+iy:x,y\in\mathbb{R}\}$$ 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 $$d((0+0i),(0+1i))=\sqrt{0+1}=1.$$

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