anonymous
  • anonymous
Show that the singular point of the following function is a pole. Determine the order m of that pole and the corresponding residue B. f(z)=(1-coshz)/(z^3)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
I really wish my university had an applied complex variables class :/
anonymous
  • anonymous
i think we can at least start this if i recall \(\cosh(z)=\cos(iz)\) if so, we can expand in a taylor series
anonymous
  • anonymous
we can try \[\cosh(z)=\sum_{k=0}^{\infty}\frac{z^{2k}}{(2k)!}\]

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anonymous
  • anonymous
this should give \[1-\cosh(z)=-\sum_{k=1}^{\infty}\frac{z^{2k}}{(2k)!}\]
anonymous
  • anonymous
yeah got it
anonymous
  • anonymous
Which would then give: \[-\sum_{k=1}^{\infty}\frac{z^{2k-3}}{(2k)!}\] right?
anonymous
  • anonymous
divide by \(z^3\) first term is \(\frac{1}{2z}\) and all the others have \(z\) in the numerator
anonymous
  • anonymous
@malevolence19 yes, that is what i get too. or shift the index either way.
anonymous
  • anonymous
So the the residue is just 1/2?
anonymous
  • anonymous
so b1 comes out to be -1/2.yeah?
anonymous
  • anonymous
yes if my memory serves me in this context the residue is the coefficient of the \(z^{-1}\) term
anonymous
  • anonymous
^right
anonymous
  • anonymous
Yeah, -1/2 i forgot about that in front.
anonymous
  • anonymous
i would not place much money on this, because it has been a while, but i am pretty sure it is correct
anonymous
  • anonymous
and pole is??i don't have any idea what the heck is it
anonymous
  • anonymous
It would just be zero right?
anonymous
  • anonymous
It it was like: \[\frac{B}{(z-\lambda)}\] then the pole would be lambda? I think, but I'm going off very little knowledge.
anonymous
  • anonymous
lambda is a singular point..not sure about pole
anonymous
  • anonymous
yes, the pole is at 0, because that is where the denominator is undefined. it is a simple pole because it is a zero of order one
anonymous
  • anonymous
^the denominator becomes undefined at singular points...so every singular point is a pole??
anonymous
  • anonymous
i think its pole is 1
anonymous
  • anonymous
a "pole" is a kind of singular point think of a simple example from pre calc class \[\frac{x^2-4}{x-2}\] vs \[\frac{x^2}{x-2}\] in the first example, 2
anonymous
  • anonymous
this thing implies m=1 at b1
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anonymous
  • anonymous
in both examples, 2 is a singular point because the function is not defined at 2 but in the first example, 2 is a removable singularity, because you can remove it
anonymous
  • anonymous
in the second example, 2 is a pole, in pre calc a "vertical asypmtote"
anonymous
  • anonymous
however it is spelled
anonymous
  • anonymous
isolated singular point.what is it??
anonymous
  • anonymous
ok hold the phone
anonymous
  • anonymous
what you posted was about the order of the pole, not the pole itself
anonymous
  • anonymous
in your case above, we had \(\frac{1}{2z}\) plus stuff with positive exponents, means you have a pole of ORDER 1, not a pole at \(z=1\)
anonymous
  • anonymous
here is a nice succinct explanation (very short) with example http://www.solitaryroad.com/c612.html
anonymous
  • anonymous
implies order of pole=k if bk is the co-efficient of z^-1 term.right?
anonymous
  • anonymous
the order of the pole is the multplicty of the zero on the denominator, to put it in pre calculus terms, example \[\frac{1}{(z-5)^3}\] has a pole of order 3 at \(z=5\)
anonymous
  • anonymous
\[\frac{1}{z-2}\] has a pole of order 1 at \(z=2\) a pole of order one is called a simple pole
anonymous
  • anonymous
oh *shouts got it*
anonymous
  • anonymous
good!
anonymous
  • anonymous
thanks bro
anonymous
  • anonymous
yw

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