## stgreen Group Title 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) one year ago one year ago

1. malevolence19

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

2. satellite73

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. satellite73

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

4. satellite73

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

5. stgreen

yeah got it

6. malevolence19

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

7. satellite73

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

8. satellite73

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

9. malevolence19

So the the residue is just 1/2?

10. stgreen

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

11. satellite73

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

12. stgreen

^right

13. malevolence19

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

14. satellite73

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

15. stgreen

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

16. malevolence19

It would just be zero right?

17. malevolence19

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. stgreen

lambda is a singular point..not sure about pole

19. satellite73

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. stgreen

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

21. stgreen

i think its pole is 1

22. satellite73

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. stgreen

this thing implies m=1 at b1

24. satellite73

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. satellite73

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

26. satellite73

however it is spelled

27. stgreen

isolated singular point.what is it??

28. satellite73

ok hold the phone

29. satellite73

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

30. satellite73

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. satellite73

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

32. stgreen

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

33. satellite73

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. satellite73

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

35. stgreen

oh *shouts got it*

36. satellite73

good!

37. stgreen

thanks bro

38. satellite73

yw