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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Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...

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