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 one year ago
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)=(1coshz)/(z^3)
 one year ago
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)=(1coshz)/(z^3)

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malevolence19
 one year ago
Best ResponseYou've already chosen the best response.0I really wish my university had an applied complex variables class :/

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0i think we can at least start this if i recall \(\cosh(z)=\cos(iz)\) if so, we can expand in a taylor series

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0we can try \[\cosh(z)=\sum_{k=0}^{\infty}\frac{z^{2k}}{(2k)!}\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0this should give \[1\cosh(z)=\sum_{k=1}^{\infty}\frac{z^{2k}}{(2k)!}\]

malevolence19
 one year ago
Best ResponseYou've already chosen the best response.0Which would then give: \[\sum_{k=1}^{\infty}\frac{z^{2k3}}{(2k)!}\] right?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0divide by \(z^3\) first term is \(\frac{1}{2z}\) and all the others have \(z\) in the numerator

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0@malevolence19 yes, that is what i get too. or shift the index either way.

malevolence19
 one year ago
Best ResponseYou've already chosen the best response.0So the the residue is just 1/2?

stgreen
 one year ago
Best ResponseYou've already chosen the best response.4so b1 comes out to be 1/2.yeah?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0yes if my memory serves me in this context the residue is the coefficient of the \(z^{1}\) term

malevolence19
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, 1/2 i forgot about that in front.

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0i would not place much money on this, because it has been a while, but i am pretty sure it is correct

stgreen
 one year ago
Best ResponseYou've already chosen the best response.4and pole is??i don't have any idea what the heck is it

malevolence19
 one year ago
Best ResponseYou've already chosen the best response.0It would just be zero right?

malevolence19
 one year ago
Best ResponseYou've already chosen the best response.0It it was like: \[\frac{B}{(z\lambda)}\] then the pole would be lambda? I think, but I'm going off very little knowledge.

stgreen
 one year ago
Best ResponseYou've already chosen the best response.4lambda is a singular point..not sure about pole

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0yes, 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

stgreen
 one year ago
Best ResponseYou've already chosen the best response.4^the denominator becomes undefined at singular points...so every singular point is a pole??

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0a "pole" is a kind of singular point think of a simple example from pre calc class \[\frac{x^24}{x2}\] vs \[\frac{x^2}{x2}\] in the first example, 2

stgreen
 one year ago
Best ResponseYou've already chosen the best response.4this thing implies m=1 at b1

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0in 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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0in the second example, 2 is a pole, in pre calc a "vertical asypmtote"

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0however it is spelled

stgreen
 one year ago
Best ResponseYou've already chosen the best response.4isolated singular point.what is it??

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0what you posted was about the order of the pole, not the pole itself

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0in 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 \(z=1\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0here is a nice succinct explanation (very short) with example http://www.solitaryroad.com/c612.html

stgreen
 one year ago
Best ResponseYou've already chosen the best response.4implies order of pole=k if bk is the coefficient of z^1 term.right?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0the order of the pole is the multplicty of the zero on the denominator, to put it in pre calculus terms, example \[\frac{1}{(z5)^3}\] has a pole of order 3 at \(z=5\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{z2}\] has a pole of order 1 at \(z=2\) a pole of order one is called a simple pole
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