Here's the question you clicked on:
stgreen
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)
I really wish my university had an applied complex variables class :/
i think we can at least start this if i recall \(\cosh(z)=\cos(iz)\) if so, we can expand in a taylor series
we can try \[\cosh(z)=\sum_{k=0}^{\infty}\frac{z^{2k}}{(2k)!}\]
this should give \[1-\cosh(z)=-\sum_{k=1}^{\infty}\frac{z^{2k}}{(2k)!}\]
Which would then give: \[-\sum_{k=1}^{\infty}\frac{z^{2k-3}}{(2k)!}\] right?
divide by \(z^3\) first term is \(\frac{1}{2z}\) and all the others have \(z\) in the numerator
@malevolence19 yes, that is what i get too. or shift the index either way.
So the the residue is just 1/2?
so b1 comes out to be -1/2.yeah?
yes if my memory serves me in this context the residue is the coefficient of the \(z^{-1}\) term
Yeah, -1/2 i forgot about that in front.
i would not place much money on this, because it has been a while, but i am pretty sure it is correct
and pole is??i don't have any idea what the heck is it
It would just be zero right?
It it was like: \[\frac{B}{(z-\lambda)}\] then the pole would be lambda? I think, but I'm going off very little knowledge.
lambda is a singular point..not sure about pole
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
^the denominator becomes undefined at singular points...so every singular point is a pole??
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
this thing implies m=1 at b1
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
in the second example, 2 is a pole, in pre calc a "vertical asypmtote"
however it is spelled
isolated singular point.what is it??
what you posted was about the order of the pole, not the pole itself
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 \(z=1\)
here is a nice succinct explanation (very short) with example http://www.solitaryroad.com/c612.html
implies order of pole=k if bk is the co-efficient of z^-1 term.right?
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 \(z=5\)
\[\frac{1}{z-2}\] has a pole of order 1 at \(z=2\) a pole of order one is called a simple pole