Show that |e^(a+bi)|=e^a and Arg(e^(a+bi))=b.
i know the first part.
= (e^a)·(e^(bi))= (e^a) ( cos b + i sin b )
=| e^(a+bi) | = (e^a)·√(cosē b + sinē b ) = e^a
I cant get the second part Arg(e^(a+bi))=b
That's because it's false if it's using the notation that I'm familiar with. It depends on whatever book you're using, but in the texts I've used, "Arg" with a capital A refers to the principle value of the polar angle - the one in between $\displaystyle - \pi$ and $\displaystyle \pi$ (I forget which endpoint is included - but only one of them is).
(Counter) Example: $\displaystyle Arg(e^{0 + 40.5 \pi i}) = Arg(e^{\frac{\pi}{2} i }) = \frac{\pi}{2}$, so $\displaystyle Arg(e^{0 + 40.5 \pi i}) \ne 40.5 \pi$.