Q:

Rotating the complex number

Suppose we have square circumscribing circle $$|z-1|=\sqrt{2}$$ and one of its vertices is $$2+\sqrt{3}i$$ so what will be other vertices ? 
I simple tried rotating half diagonal of square in anticlockwise direction , supposing we take center of circle as $z_o$ and known vertice as $z_a$ and other one as $z_b$ so 
$$\frac{z_a - z_o}{|z_a-z_o|}.e^{i.\frac{\pi}{2}} = \frac{z_b- z_o}{|z_b-z_o|}$$ which gave me 
$$( z_a - z_o).i= z_b- z_o$$ 
Now plugging in values gives me 
$$ (2+\sqrt{3}i -1).i +1= z_b $$ and finally $$1+i -\sqrt{3}$$ but answer is given as $$i -\sqrt{3}$$ , what is wrong here?? 

A:

First of all, trust your solution.  The given solution is obviously wrong because its distance from the center of the circle is different compared to $z_a$.
You don't need to normalize anything.  Just rotate around the center of the circle.  The solution is simply to move to the origin, multiply, and move back.  Let
$$z_a=2+\sqrt{3}i$$
and
$$z_0=1$$
All four vertices are simply
$$(z_a-z_0)\cdot\{1,i,-1,-i\}+z_0$$
which yields
$$2+\sqrt3 i,1+i-\sqrt3,-\sqrt3 i,1-i+\sqrt3$$
the second is your solution.