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 and (I forget which endpoint is included - but only one of them is).
(Counter) Example: , so .
Last edited by johnsomeone; Sep 25th 2012 at 02:15 PM.