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)

- anonymous

- chestercat

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- anonymous

I really wish my university had an applied complex variables class :/

- 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

we can try
\[\cosh(z)=\sum_{k=0}^{\infty}\frac{z^{2k}}{(2k)!}\]

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- anonymous

this should give
\[1-\cosh(z)=-\sum_{k=1}^{\infty}\frac{z^{2k}}{(2k)!}\]

- anonymous

yeah got it

- anonymous

Which would then give:
\[-\sum_{k=1}^{\infty}\frac{z^{2k-3}}{(2k)!}\] right?

- anonymous

divide by \(z^3\) first term is \(\frac{1}{2z}\) and all the others have \(z\) in the numerator

- anonymous

@malevolence19 yes, that is what i get too. or shift the index either way.

- anonymous

So the the residue is just 1/2?

- anonymous

so b1 comes out to be -1/2.yeah?

- anonymous

yes if my memory serves me in this context the residue is the coefficient of the \(z^{-1}\) term

- anonymous

^right

- anonymous

Yeah, -1/2 i forgot about that in front.

- anonymous

i would not place much money on this, because it has been a while, but i am pretty sure it is correct

- anonymous

and pole is??i don't have any idea what the heck is it

- anonymous

It would just be zero right?

- 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

lambda is a singular point..not sure about pole

- 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

^the denominator becomes undefined at singular points...so every singular point is a pole??

- anonymous

i think its pole is 1

- 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

this thing implies m=1 at b1

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- 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

in the second example, 2 is a pole, in pre calc a "vertical asypmtote"

- anonymous

however it is spelled

- anonymous

isolated singular point.what is it??

- anonymous

ok hold the phone

- anonymous

what you posted was about the order of the pole, not the pole itself

- 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

here is a nice succinct explanation (very short) with example
http://www.solitaryroad.com/c612.html

- anonymous

implies order of pole=k if bk is the co-efficient of z^-1 term.right?

- 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

\[\frac{1}{z-2}\] has a pole of order 1 at \(z=2\)
a pole of order one is called a simple pole

- anonymous

oh *shouts got it*

- anonymous

good!

- anonymous

thanks bro

- anonymous

yw

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