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

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- Sep 25th 2012, 12:46 PMilovemymathNeed 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)·√(cosē b + sinē b ) = e^a

I cant get the second part Arg(e^(a+bi))=b - Sep 25th 2012, 01:11 PMjohnsomeoneRe: Need help solving this. Thank you.
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 $\displaystyle Arg(e^{0 + 40.5 \pi i}) \ne 40.5 \pi$.