Question

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)

Answer 1

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

Answer 2

i think we can at least start this
if i recall \(\cosh(z)=\cos(iz)\)
if so, we can expand in a taylor series

Answer 3

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

Answer 4

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

Answer 5

yeah got it

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

So the the residue is just 1/2?

Answer 10

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

Answer 11

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

Answer 12

^right

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

It would just be zero right?

Answer 17

It it was like:
\[\frac{B}{(z-\lambda)}\] then the pole would be lambda? I think, but I'm going off very little knowledge.

Answer 18

lambda is a singular point..not sure about pole

Answer 19

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

Answer 20

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

Answer 21

i think its pole is 1

Answer 22

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

Answer 23

this thing implies m=1 at b1

Answer 24

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

Answer 25

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

Answer 26

however it is spelled

Answer 27

isolated singular point.what is it??

Answer 28

ok hold the phone

Answer 29

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

Answer 30

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 \(z=1\)

Answer 31

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

Answer 32

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

Answer 33

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 \(z=5\)

Answer 34

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

Answer 35

oh *shouts got it*

Answer 36

good!

Answer 37

thanks bro

Answer 38

yw