I must be true. For any real number $\alpha$,
$\begin{array}{cl}
& u \\
= & w +\dfrac{1}{w} \\
= & \dfrac{\alpha +i}{\alpha -i} +\dfrac{\alpha -i}{\alpha +i} \\
= & \dfrac{(\alpha+i)^2 +(\alpha-i)^2}{(\alpha -i)(\alpha +i)} \\
= & \dfrac{\alpha^2 +2\alpha i +i^2 +\alpha^2-2\alpha i +i^2}{\alpha^2 -i^2} \\
= & \dfrac{2\alpha^2 +2i^2}{\alpha^2-i^2} \\
= & \dfrac{2\alpha^2 +2(-1)}{\alpha^2-(-1)} \\
= & \dfrac{2(\alpha^2-1)}{\alpha^2+1} \text{, which is independent of $i$.}
\end{array}$
II must be true. For any real number $\alpha$,
$\begin{array}{cl}
& v \\
= & w -\dfrac{1}{w} \\
= & \dfrac{\alpha +i}{\alpha -i} -\dfrac{\alpha -i}{\alpha +i} \\
= & \dfrac{(\alpha+i)^2 -(\alpha-i)^2}{(\alpha -i)(\alpha +i)} \\
= & \dfrac{\alpha^2 +2\alpha i +i^2 -\alpha^2+2\alpha I -i^2}{\alpha^2 -i^2} \\
= & \dfrac{4\alpha i}{\alpha^2-i^2} \\
= & \dfrac{4\alpha i}{\alpha^2-(-1)} \\
= & \dfrac{4\alpha}{\alpha^2+1}i
\end{array}$
Therefore, the real part of $v$ is $0$.
III may not be true.
$\begin{array}{cl}
& w \\
= & \dfrac{\alpha +i}{\alpha -i} \\
= & \dfrac{\alpha +i}{\alpha -I} \times \dfrac{\alpha +i}{\alpha +i} \\
= & \dfrac{(\alpha +i)^2}{\alpha^2 -i^2} \\
= & \dfrac{\alpha^2 +2\alpha i +i^2}{\alpha^2 -i^2} \\
= & \dfrac{\alpha^2 +2\alpha i +(-1)}{\alpha^2 -(-1)} \\
= & \dfrac{\alpha^2 -1}{\alpha^2 +1} + \dfrac{2\alpha i}{\alpha^2 +1} \\ \\
\end{array}$
Also,
$\begin{array}{cl}
& 2w \\
= & \dfrac{2(\alpha^2 -1)}{\alpha^2 +1} + \dfrac{4\alpha i}{\alpha^2 +1} \\
\end{array}$
Obviously, it is not a must that $\dfrac{2\alpha i}{\alpha^2 +1} = \dfrac{4\alpha i}{\alpha^2 +1}$.
