I am trying to solve $$e^{3z} = 1+i$$
Putting the RHS into modulus argument form, $$1+i = \sqrt{2} e^{(\pi/4 +2\pi n)i}$$
Now what I want to do is equate moduli and arguments, letting $z = a+bi$
$$e^{3a+3bi} = \sqrt{2} e^{(\pi/4 +2\pi n)i}$$
So then can I say that $$e^{3a} = \sqrt{2} \quad,\quad 3b = \pi/4 +2\pi n$$
This would then finish the problem, but I am not sure if this is allowed. Thanks a lot.
You can make more clear what you're doing this way;
$$e^{3a+3bi}=\sqrt 2 e^{(\pi/4+2\pi n)i} \iff e^{3a+3bi}=e^{(\pi/4+2\pi n)i+\log (\sqrt 2)} \stackrel{\log}{\iff} 3a+3bi=\left (\frac{\pi}{4}+2\pi n\right )i+\log \left (\sqrt 2\right ) \iff 3a+3bi=\frac12\log 2+\left (\frac{\pi}{4}+2\pi n\right )i$$