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?
 A: 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.
A: Note that $i=(0,1)$ and the origin is $(0,0)$
According to the distance formula we get $d= \sqrt {0^2+1^2} =1$
A: 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.
A: 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 $|i|=\sqrt 1$.
A: $\mathbb{C}=\{x+iy:x,y\in\mathbb{R}\}$ 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 $d((0+0i),(0+1i))=\sqrt{0+1}=1.$
A: We simply have by definition
$$|i|=\sqrt{1^2}=1$$

(credit Wikipedia)
