Use of Norton theorem to find short circuit current
We're using superposition to establish the short circuit curent. When ignoring \$i_1\$ we can calculate the impedance between terminals A and B:

$$ Z_{AB} = j \omega L_2 + \frac{1}{j \omega C + 1/R} $$

leading to a current of
$$ i_{u1} = \frac{u_1}{Z_{AB}} = \frac{u_1}{j \omega L_2 + \frac{1}{j \omega C + 1/R}}$$

This is the first partial current. Now, if we consider \$i_1\$ and replace \$u_1\$ with a short circuit, \$i_1\$ is connected directly between terminals A and B. Thus:

$$ i_{short} = i_1 + i_{u1} $$

So you are half correct. The short circuit current consists of a part \$i_1\$ and a part \$i_{u1}\$ contributed by the voltage source.