To solve this problem, we need to analyze a lossless transmission line terminated with a short circuit, and calculate the normalized impedance at different points as we move from the load towards the generator.

For a lossless transmission line terminated with a short circuit, we know that:

1. The input impedance of a short-circuited transmission line of length \(\lambda/8\) is given by:

\(Z_{in} = jZ_0 \tan(\beta \ell)\)

where \(\beta = \frac{2\pi}{\lambda}\) is the phase constant, \(\ell\) is the length of the line, and \(Z_0\) is the characteristic impedance.

2. Since the line is short-circuited:

At \(\ell = \lambda/8\), \(\tan(\beta \ell) = \tan(\pi/4) = 1\), so \(Z_{in} = jZ_0\)

At \(\ell = \lambda/4\), \(\tan(\beta \ell) = \tan(\pi/2) = \infty\), so \(Z_{in} = \infty\)

At \(\ell = 3\lambda/8\), \(\tan(\beta \ell) = \tan(3\pi/4) = -1\), so \(Z_{in} = -jZ_0\)

At \(\ell = \lambda/2\), \(\tan(\beta \ell) = \tan(\pi) = 0\), so \(Z_{in} = 0\)

Let's match these results with the points:

• \(Z_{inA} = +j\Omega\) (at \(\lambda/8\))
• \(Z_{inB} = \infty\) (at \(\lambda/4\))
• \(Z_{inC} = -j\Omega\) (at \(3\lambda/8\))
• \(Z_{inD} = 0\) (at \(\lambda/2\))

Hence, the correct answer is $Z_{inA}=+1j\Omega$, $Z_{inB}=\infty$, $Z_{inC}=-1j\Omega$, $Z_{inD}=0$.