# Need help solving this. Thank you.

• Sep 25th 2012, 01:46 PM
ilovemymath
Need help solving this. Thank you.
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)·√(cos² b + sin² b ) = e^a

I cant get the second part
Arg(e^(a+bi))=b
• Sep 25th 2012, 02:11 PM
johnsomeone
Re: Need help solving this. Thank you.
Quote:

Originally Posted by ilovemymath
Show that |e^(a+bi)|=e^a and Arg(e^(a+bi))=b.
...
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 $- \pi$ and $\pi$ (I forget which endpoint is included - but only one of them is).

(Counter) Example: $Arg(e^{0 + 40.5 \pi i}) = Arg(e^{\frac{\pi}{2} i }) = \frac{\pi}{2}$, so $Arg(e^{0 + 40.5 \pi i}) \ne 40.5 \pi$.