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## anonymous 3 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)=(1-coshz)/(z^3)

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1. anonymous

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

2. 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

3. anonymous

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

4. anonymous

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

5. anonymous

yeah got it

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

So the the residue is just 1/2?

10. anonymous

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

11. anonymous

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

12. anonymous

^right

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

It would just be zero right?

17. 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.

18. anonymous

lambda is a singular point..not sure about pole

19. 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

20. anonymous

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

21. anonymous

i think its pole is 1

22. 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

23. anonymous

this thing implies m=1 at b1

24. 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

25. anonymous

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

26. anonymous

however it is spelled

27. anonymous

isolated singular point.what is it??

28. anonymous

ok hold the phone

29. anonymous

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

30. 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 $$z=1$$

31. anonymous

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

32. anonymous

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

33. 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 $$z=5$$

34. anonymous

$\frac{1}{z-2}$ has a pole of order 1 at $$z=2$$ a pole of order one is called a simple pole

35. anonymous

oh *shouts got it*

36. anonymous

good!

37. anonymous

thanks bro

38. anonymous

yw

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