(c)

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{F(j)=qj{q}^{-1}}& =(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)(j)(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)\\ \multicolumn{1}{c}{}& =\frac{1}{2}(1+i)(j)((1-i)\\ \multicolumn{1}{c}{}& =\frac{1}{2}(j+k)(1-i)=\frac{1}{2}(j+k+k-j)\\ \multicolumn{1}{c}{}& =k\end{array}.}$

Similarly $Fk=-j$. $F$ is a rotation of 90 degrees about the x axis.

(2)

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{qv{q}^{-1}}& =(\cos \theta +\sin \theta k)(r\cos \varphi i+\sin \varphi j)(\cos \theta -\sin \theta k)\\ \multicolumn{1}{c}{}& =r((\cos \theta \cos \varphi -\sin \theta \sin \varphi )i+(\cos \theta \sin \varphi +\sin \theta \cos \varphi )j)(\cos \theta i-\sin \theta k)\\ \multicolumn{1}{c}{}& =r(\cos (\theta +\varphi )i+\sin (\theta +\varphi )j)(\cos \theta i-\sin \theta k)\\ \multicolumn{1}{c}{}& =r((\cos (\theta +\varphi )\cos \theta -\sin (\theta +\varphi )\sin \varphi )i+(\cos (\theta +\varphi )\sin \theta +\sin (\theta +\varphi )\cos \varphi )j)\\ \multicolumn{1}{c}{}& =r(\cos (2\theta +\varphi )i+\sin (2\theta +\varphi )j)\end{array}.}$