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anonymous
 4 years 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)
anonymous
 4 years 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I really wish my university had an applied complex variables class :/

anonymous
 4 years 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

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

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

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

anonymous
 4 years 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So the the residue is just 1/2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so b1 comes out to be 1/2.yeah?

anonymous
 4 years 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, 1/2 i forgot about that in front.

anonymous
 4 years 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and pole is??i don't have any idea what the heck is it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It would just be zero right?

anonymous
 4 years 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lambda is a singular point..not sure about pole

anonymous
 4 years 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0^the denominator becomes undefined at singular points...so every singular point is a pole??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think its pole is 1

anonymous
 4 years 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this thing implies m=1 at b1

anonymous
 4 years 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0however it is spelled

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0isolated singular point.what is it??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what you posted was about the order of the pole, not the pole itself

anonymous
 4 years 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 \(z=1\)

anonymous
 4 years 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0implies order of pole=k if bk is the coefficient of z^1 term.right?

anonymous
 4 years 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 \(z=5\)

anonymous
 4 years 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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