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

liyz06
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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

 
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Let z=x+iy. Suppose you take the limit along the line x=0. What happens?
 
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 z=\rho e ^{i\theta} \Rightarrow e^{-z^2}=e^{-\rho^2e^{2i\theta}}, the magnitude is really dependent on Re(e^{2i\theta})=\cos 2\theta&gt;0, and that's where the |arg(z)|&lt;\pi/4 from
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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