How would one find the fixed points of $e^z$, where $z$ is complex (if there are any)? I feel this problem probably has a really obvious answer, and for some reason, I'm just not getting it. Thanks.
Fixed points of $e^z$
1
$\begingroup$
functions
complex-numbers
exponentiation
-
0You could start with the definition of e^z when z=x+iy. – 2012-01-05
-
0Here's a bit more...$e^{x+iy}=e^x(\cos(y)+i\sin(y))$ so $|e^{x+iy}|=e^x$. So if $e^{x+iy}=x+iy$ you'll need to have $e^x=|x+iy|=x^2+y^2$ (among other things). – 2012-01-05
-
0Ok. I would write e^z = e^x * e^iy = e^x * (cos(y) + isin(y))... but then how would I proceed. I've tried manipulating it and it doesn't seem to work. Also, I realised that I need sqrt(x^2 + y^2) = e^x, and y = arctan(y/x).. – 2012-01-05