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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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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