Why Does exp(-z^2) Approach Zero in Certain Sectors?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
liyz06
Messages
6
Reaction score
0

Homework Statement



Reading Hinch's book, there is a statement as follows:

... z need to be kept in the sector where exp(-z^2) ->0 as z -> infinity. Thus it's applicable to the sector |arg z|<pi/4...

Homework Equations



Why is this true and what is the limiting behavior of exp(x) for x in different sectors of the complex plane?

The Attempt at a Solution

 
Physics news on Phys.org
vela said:
Let z=x+iy. Suppose you take the limit along the line x=0. What happens?

I get it now, use polar coordinate then it's [itex]z=\rho e ^{i\theta} \Rightarrow e^{-z^2}=e^{-\rho^2e^{2i\theta}}[/itex], the magnitude is really dependent on [itex]Re(e^{2i\theta})=\cos 2\theta>0[/itex], and that's where the [itex]|arg(z)|<\pi/4[/itex] from
 
Last edited: