geometry - How to find vertices of square in complex numbers

Given a square with vertices: $\sqrt3+i,-1+i\sqrt3,-i-\sqrt3,1-i\sqrt3$

So that a circle that is centered in the origin is blocking the square.

My question is how can I find the vertices of a square (different one) that blocks this origin centerd circle such that his sides are parallel to the first square I mentioned.

Is there any easy way to solve it?

Thanks.

Answer

You started with a square $Q$, centered at the origin, whose diagonals have length $d$. The circle $C$ circumscribing this square has diameter $d$. You want to find the square $Q'$ circumscribing $C$ whose sides are parallel to the sides of $Q$.

Since $Q$ is centered at the origin, $Q'$ is just a rescaling of $Q$. In other words, there is some positive real number $\lambda$ with the property that $Q' = \{\lambda z : z \in Q\}$. For short, I'll just write $Q' = \lambda Q$.

To get $Q'$, all we have to do is find $\lambda$. To find $\lambda$, notice that the sides of $Q'$ must have length $d$. On the other hand, the sides of $Q$ have length $\frac{d}{\sqrt{2}}$, so the sides of $Q'$ will have length $\lambda \frac{d}{\sqrt{2}}$. That means you should pick $\lambda = \sqrt{2}$, just like you guessed.

@Semiclassical drew a nice picture of $Q$, $C$, and $Q'$. Looking at the right triangles formed by the extra lines they drew from the center, you can see quite directly that the diagonals of $Q$, the diameter of $C$, and the sides of $Q'$ are the same length.