Question

find the residue of the following function at the point 0
f(z)=(1+z^2)/z
can you tell me how its worked out, please

Answer 1

at z=0 function has a simple pole, so
residue will be 1

Answer 2

can you show me what the laurent series will be of this, please?

Answer 3

f(z) can be written like:
f(z)=1/z +z
and this is it's Laurent series

Answer 4

Ok?

Answer 5

thanks.i understand.
my book essentially says the same thing but it shows the following..
(1+z^2)/z =..+0/z^3 +0/z^2 + 1/z +0 +z +0z^2...

I am not understanding why they are including all the other bits inbetween and how and why they come about... could oyu explain please

Answer 6

I think this is by definition:
\[a_n=\frac{1}{2\pi i}\int\limits_\gamma\frac{f(z)}{(z-c)^{n+1}}\]

Answer 7

so for the same function in the question..for alpha=0..how would i work out \[\alpha _{-1}/(z-\alpha\])

Answer 8

but your function is the case where \(\alpha=0 \)

Answer 9

ok ignoring alpha =0

Answer 10

Or I am not understanding right your question....

Answer 11

how would i work that
α−1/(z−α
)

Answer 12

what you need?

Answer 13

according to my book..
the residue of an analytic function f at a singularity α of f is the coeffiecient \[\alpha _{-1}\] of \[(z-\alpha)^{-1}\] in the laurent series for f :
f(z)=\[...+\alpha _{-2}/(z-\alpha) +\alpha_{-1}/(z-\alpha) +\alpha_{0} +\alpha_{1}(z-\alpha)+...\], we can determine Res(f,alpha) by inspecting the series..

now my question specifically is how would i work out the coeffecient \[\alpha _{-1}\]

Answer 14

By finding the Laurent seies, :)

Answer 15

and that is by using the definition you stated a few posts up right?

Answer 16

there are also two formulas:
http://en.wikipedia.org/wiki/Residue_(complex_analysis)#Limit_formula_for_higher_order_poles

Answer 17

for simple and higher order poles. Or ya, by using formula I gave you above. But it can be tough to calculate the actual integral. Normaly finding Laurent series is easyer

Answer 18

aaargghh.. this is really chewing my brain out...
I think i get it...i'll have to attempt a few questions and maybe do some more reading to get my head around it.
Thanks Myko..you are a star for helping. wish i could give you more medals..

Answer 19

:). Good luck. Complex analysis is tough at the beginning