I am having troubles with finding and argument of these two $$\frac{1}{i}, \frac{2^{e^{i \theta}}}i $$
for the first one my approach was
$$|z|=\frac{1}ie^0$$
$$e^{i\theta}=e^0$$
$$\theta=0$$
$$|z|=\frac{1}i(\cos\theta+i\sin\theta)$$
I'm stuck here :(
for the second one I've no idea from where to start :(
I guess you should use Euler's identity. $$e^{i\theta} = \cos \theta+i \sin \theta$$
The first one $\frac{i}{1} = i = e^{i(\frac{\pi}{2}+2k\pi)}$
The second is more interesting because of the numerator:
$$2^{e^{i\theta}}= 2^{\cos \theta + i\sin \theta} = 2^{\cos \theta}\cdot 2^{i \sin \theta}$$
It only remains to be put in base $e$, you can do this as follows:
$$2^{i\sin \theta} = e^{\ln(2^{i\sin \theta})} = e^{i\sin \theta \cdot \ln 2}$$
You just need to combine these different steps in order to get the argument.