## 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 Group Title

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

2. satellite73 Group Title

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 Group Title

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

4. satellite73 Group Title

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

5. stgreen Group Title

yeah got it

6. malevolence19 Group Title

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

7. satellite73 Group Title

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

8. satellite73 Group Title

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

9. malevolence19 Group Title

So the the residue is just 1/2?

10. stgreen Group Title

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

11. satellite73 Group Title

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

12. stgreen Group Title

^right

13. malevolence19 Group Title

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

14. satellite73 Group Title

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

15. stgreen Group Title

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

16. malevolence19 Group Title

It would just be zero right?

17. malevolence19 Group Title

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 Group Title

lambda is a singular point..not sure about pole

19. satellite73 Group Title

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 Group Title

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

21. stgreen Group Title

i think its pole is 1

22. satellite73 Group Title

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 Group Title

this thing implies m=1 at b1

24. satellite73 Group Title

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 Group Title

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

26. satellite73 Group Title

however it is spelled

27. stgreen Group Title

isolated singular point.what is it??

28. satellite73 Group Title

ok hold the phone

29. satellite73 Group Title

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

30. satellite73 Group Title

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 Group Title

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

32. stgreen Group Title

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

33. satellite73 Group Title

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 Group Title

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

35. stgreen Group Title

oh *shouts got it*

36. satellite73 Group Title

good!

37. stgreen Group Title

thanks bro

38. satellite73 Group Title

yw