Is $i$ a rotation? 
Question: Is $i$ (the imaginary number) a rotation?


This might be a very a loose question to ask but I can't find a better place to ask. Now a friend of mine told me this when we were discussing complex numbers, but I just can't figure out how is it a rotation? I know a complex numbers do work somewhat like vectors which deals with a lot of rotations and stuff, but I can't connect that with $i$ being a rotation. Now it can just be false in case he lied, but if it's true then please I want some explanation on this. Thanks a lot.
 A: The imaginary number $i$ is not a rotation.
The multiplication by $i$, so the map $z\mapsto z\cdot i$, can be seen as a rotation of the plane $\mathbb C$
A: Your friend did not really lie, but instead used a shortcut.
It's similar to mathematical properties that we learn about addition and multiplication of real numbers. For example, think about the number line.

Addition of $1$ is a translation of the real number line, meaning that simultaneously adding $1$ to every point on the number line is to translate (or slide) the entire number to the right by 2 units.

Also,

Multiplication by $2$ is an expansion of the real number line, meaning that simultaneously multiplying every point on the number line by $2$, has the geometrical effect of stretching the entire number line by a factor of $2$.

Now let's turn to complex numbers. I'm sure you have already learned that complex numbers $a+bi$ can be plotted in the Cartesian coordinate plane: $a+bi$ is plotted as $(a,b)$. Consider now the complex number $i = 0+1i$. What your friend said, without any shortcut, is this:

Multiplication by $i$ is a rotation of the plane through an angle of $90^\circ$, meaning that simultaneously multiplying the entire complex plane by $i$ has the geometrical effect of rotating the entire plane around the origin by a positive angle of $90^\circ$.

I'm sure you can verify this for yourself with some simple computations. For example, you can easily compute $(1+2i) \cdot i = -2+i$; and you can easily verify that the angle from the vector $(1,2)$ to the vector $(-2,1)$ is $90^\circ$ in the counterclockwise direction. More generally, for any complex number $a+bi$ you can compute $(a+bi) \cdot i = -b + ai$, and you can easily verify that the angle from the vector $(a,b)$ to the vector $(-b,a)$ is $90^\circ$.
