What's the correct method to find the Thevenin Impedance in this circuit?
You could add a test source at the output instead of trying to solve by means of the short circuit current. 

For the test source method, you have to turn off the independent sources. So your circuit will look like this: 







simulate this circuit &ndash; Schematic created using CircuitLab

Your Thevenin impedance is \$Z_{\mathrm{TH}}=\dfrac{V_t}{i_t}\$, where \$V_t\$ and \$i_t\$ are your test source and current source respectively. 

Using KCL:

$$i_a+0.2V_o =i_t$$

$$\dfrac{0-V_t}{4\Omega-j2\Omega+8\Omega+j4\Omega}+0.2(8\Omega+j4\Omega)\bigg(\dfrac{0-V_t}{4\Omega-j2\Omega+8\Omega+j4\Omega}\bigg)=i_t $$

$$ \dfrac{-V_t}{12\Omega+j2\Omega}-0.2(8\Omega+j4\Omega)\bigg(\dfrac{V_t}{12\Omega+j2\Omega}\bigg)=i_t$$

$$ \dfrac{V_t}{12\Omega+j2\Omega}\bigg[-1-0.2(8\Omega+j4\Omega)\bigg]=i_t$$

$$ \dfrac{V_t}{12\Omega+j2\Omega}\bigg[-2.6\Omega-j0.8\Omega\bigg]=i_t$$
$$ \dfrac{V_t}{i_t}=\dfrac{12\Omega+j2\Omega}{-2.6\Omega-j0.8\Omega}$$

$$ \dfrac{V_t}{i_t}=Z_{\mathrm{TH}}=4.47\angle{172.36^{\mathrm{o}}}$$

Now, this answer is different from the one your are given. The only way I can see you getting the \$12.166\angle{136.6^\mathrm{o}}\$ answer is if you reversed the polarity of \$V_o\$, I am very confident that's where the sign affect their answer.

I see in one of your comments that you tried a circuit solver and got the following for \$i_{sc}\$

$$ i_{sc}=-0.27-j1.62=1.64\angle{-99.46}$$

This could be used to validate the answer I obtained:

$$Z_{\mathrm{TH}}=\dfrac{V_{\mathrm{TH}}}{i_{sc}}=\dfrac{7.35\angle{72.9}}{1.64\angle{-80.54}}=4.48\angle{172.36}$$

That should be the answer.