Rotating the complex number

Question

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??

Answer 1

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.