If $x$ in a division $\mathbb{R}$-algebra and $x(a+bi)x^{-1}=a-bi, \forall a,b \in \mathbb{R}$, show that $x^2$ commutes with all $z \in \mathbb{C}$.

Question

If $x$ in a division $\mathbb{R}$-algebra and $x(a+bi)x^{-1}=a-bi, \forall a,b \in \mathbb{R}$, show that $x^2$ commutes with all $z \in \mathbb{C}$.

I have tried squaring both sides of $x(a+bi)x^{-1}=a-bi,$ but this doesn't help. Do I need to find a special expression for $z$? I feel like this should be really easy, but can't find the way to do it.

Answer 1

For all $a,b\in\mathbb{R}$, we have $x^2(a+bi)x^{-2}=x(x(a+bi)x^{-1})x^{-1}=x(a-bi)x^{-1}=a+bi$ (apply your property replacing $b$ by $-b$).

Hence, $x^2$ commutes with $a+bi$ for all $a,b\in\mathbb{R}$, as required.