For a lossless transmission line of characteristic impedance \(Z_0\) and length \(l\), terminated by a load \(Z_L\), the input impedance \(Z_{in}\) is given by:
\[ Z_{in} = Z_0 \frac{Z_L + jZ_0 \tan(\beta l)}{Z_0 + jZ_L \tan(\beta l)} \]
where \(\beta = 2\pi/\lambda\).
Let's analyze each statement for a lossless line:
(a) Shorted line (\(Z_L = 0\)), \(l = \lambda/8\).
\(\beta l = (2\pi/\lambda) (\lambda/8) = \pi/4\). So \(\tan(\beta l) = \tan(\pi/4) = 1\).
\(Z_{in} = Z_0 \frac{0 + jZ_0(1)}{Z_0 + j(0)(1)} = Z_0 \frac{jZ_0}{Z_0} = jZ_0\).
The statement says \(Z_{in} = -jZ_0\). This is FALSE. \(Z_{in}\) should be \(+jZ_0\).
So, this statement is "not true".
(b) Shorted line (\(Z_L = 0\)), \(l = \lambda/4\).
\(\beta l = (2\pi/\lambda) (\lambda/4) = \pi/2\). So \(\tan(\beta l) = \tan(\pi/2) = \infty\).
For this case, it's easier to use \(Z_{in} = jZ_0 \tan(\beta l)\) for a shorted line.
\(Z_{in} = jZ_0 \tan(\pi/2) = jZ_0 (\infty) = j\infty\). This represents an open circuit.
The statement says \(Z_{in} = j\infty\). This is TRUE.
(c) Matched line (\(Z_L = Z_0\)).
\(Z_{in} = Z_0 \frac{Z_0 + jZ_0 \tan(\beta l)}{Z_0 + jZ_0 \tan(\beta l)} = Z_0 \frac{Z_0(1 + j\tan(\beta l))}{Z_0(1 + j\tan(\beta l))} = Z_0\).
The statement says \(Z_{in} = Z_0\). This is TRUE.
(d) At a half-wavelength from load (\(l = \lambda/2\)), \(Z_{in} = Z_L\), and repeats for every half wavelength thereafter.
If \(l = n\lambda/2\) (where n is an integer), then \(\beta l = (2\pi/\lambda) (n\lambda/2) = n\pi\).
\(\tan(\beta l) = \tan(n\pi) = 0\).
Then \(Z_{in} = Z_0 \frac{Z_L + jZ_0(0)}{Z_0 + jZ_L(0)} = Z_0 \frac{Z_L}{Z_0} = Z_L\).
This is TRUE. The input impedance repeats every half-wavelength.
Therefore, the statement that is "not true" is (a).
\[ \boxed{Z_{in} = -jZ_0 \text{ for a shorted line with } l = \frac{\lambda}{8}} \]