How to translate a point by $90$ degree counter clockwise direction? I would like to find a point which is rotated $90$ degree counter clockwise direction about the origin. For example, the point $(2,0)$ is taken to $(0,2)$.  
Given $(x,y)$, how do i find the new point which is rotated $90$ degree counter clockwise  direction.
My approach:
From the given $(x,y)$, we can calculate its distance from origin. i.e $r=\sqrt{(x^2+y^2)}$. From polar co-ordinates, we can find the angle i.e $\sin \theta = \frac{y}{r} \text{ or } \cos \theta = \frac{x}{r}$.  
Once we know $(r,\theta)$, now the translated point as $(r\cos(\theta+90),r\sin(\theta+90))$
Is there any other shorted method to find the translation?
 A: For counter clockwise roation you can multply the point with the roation matrix
$$\begin{pmatrix} \cos(\theta) & -\sin(\theta)\\\sin(\theta) & \cos(\theta)\\ \end{pmatrix}$$
where $\theta$ is the degree. 
For $90$ degree we have 
$$\begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}$$
So that would be 
$$\begin{pmatrix} 0 & - 1\\ 1 & 0 \\ \end{pmatrix}\cdot \begin{pmatrix} x \\ y \\\end{pmatrix}= \begin{pmatrix} -y \\ x \end{pmatrix}$$
The idea is that roation is a linear operator and those can be written as a matrix. 
If you rotate the point $(1,0)$  90 degrees you get $(0,1)$ and if you rotate $(0,1)$ you get $(-1,0)$. if you check the behavior of those points when you multply with the matrix you find out that it is the same.
A: For a $\frac{\pi}{2}$ anticlockwise rotation of $(x,y)$ about the origin, you can just take $(-y,x)$. This is about as short as you can get.
To see why, note that $\cos(\theta+\frac{\pi}{2}) = -\sin \theta$, and $\sin(\theta+\frac{\pi}{2}) = \cos \theta$. Hence the point $(r \cos \theta, r \sin \theta)$ will become $(r \cos (\theta+\frac{\pi}{2}), r \sin (\theta+\frac{\pi}{2}))$ = $(-r \sin \theta, r \cos \theta)$. Comparing coordinates gives the formula above.
A: $x=rcos\theta, iy=risin\theta$
$e^\theta=cos\theta +isin \theta$
On an argand plane, any point is represented by $re^\theta$.
If you have to have to rotate the angle by $\frac{\pi}{2}$ .
We have to obtain this: $rcos(\theta+\frac {\pi}{2})+r isin(\theta+\frac {\pi}{2})$
$e^\frac {\pi}{2}= cos \frac{\pi}{2} + isin\frac{\pi}{2}=i$
So you just have to multiply $e^\theta$ by $e^\frac {\pi}{2}$.
Which gives $icos\theta+ i^2sin\theta=icos\theta-sin\theta$. 
$icos\theta=isin(\theta+\frac {\pi}{2})$
$cos(\theta+\frac {\pi}{2})=-sin\theta$.
Therefore, when you rotate by $\frac{\pi}{2}$ , the co-ordinates are $(-y,x)$.
You can really try multiplying other roots and powers of $i$ to rotate the co-ordinates by standard angles. :) 
