It's actually quite straightforward to find the Thevenin impedance of this circuit.

The equivalent impedance looking into the port ab is defined by:

$$Z_{ab} \equiv \frac{V_{ab}}{I_{ab}} = \frac{V_{ab}}{i_{\Delta}} = Z_{th}$$

But you can write by inspection a simple KVL equation for \$V_{ab}\$ in terms of \$i_{\Delta}\$:

$$V_{ab} = i_{\Delta}(-jX_{1uF} + 10k\Omega) + 200 i_{\Delta} \cdot 100 \Omega) = i_{\Delta}(-jX_{1uF} + 30k)\Omega$$

It's actually quite straightforward to find the Thevenin impedance of this circuit.

The equivalent impedance looking into the port ab is defined by:

$$Z_{ab} \equiv \frac{V_{ab}}{I_{ab}} = \frac{V_{ab}}{i_{\Delta}} = Z_{th}$$

But you can write by inspection a simple KVL equation for \$V_{ab}\$ in terms of \$i_{\Delta}\$:

$$V_{ab} = i_{\Delta}(-jX_{1uF} + 10k\Omega) + 200 i_{\Delta} \cdot 100 \Omega) = i_{\Delta}(-jX_{1uF} + 30k)\Omega$$

Generally speaking, to find the Thevenin equivalent of a circuit with only dependent source(s), you must be sure to "activate" the dependent source(s) with a test source.

This is what was done above. We solved for the voltage across the port due to a test current source, \$I_{ab}\$ which, in this case, equals the controlling variable thus making this problem particularly easy to solve.