Question

ok i am still stuck. expand
\[\frac{z^2-1}{(z^2+1)^2}\] at
\[z_0=i\]

Answer 1

jamesj suggesting rewriting as
\[\frac{1}{(z-i)^2}\frac{z^2-1}{(z+i)^2}\] but the second part is killing me

Answer 2

Satellite I didn't know u ever needed help

Answer 3

for some reason this laurant expansion is kicking my butt. i know the answer, but i don't know a good way to get it

Answer 4

I can't even figure out how jamesj got that!!!!!!

Answer 5

denominator is
\[(z^2+1)^2=(z+i)(z-i)^2=(z-i)^2(z+i)^2\] is how,but the expansion is eluding me

Answer 6

That isn't the right factorization of the denominator. You used difference of squares, but a sum of squares cannot be factored.

Answer 7

@jamesj i tried what you suggested, but it is not clear to me how dividing
\[\frac{z^2-1}{(z+i)^2}\] get me the expansion. i just get a quotient of 1 and a remainder of something, i forget what

Answer 8

Satellite -- I'm at work right now so I don't have much time, but if you haven't already, you should check out this free online set of notes about complex analysis. It's rigorous and readable. The guy taught it for 20 years, so he knows his stuff :

http://rutherglen.science.mq.edu.au/wchen/lnicafolder/lnica.html

Hope this helps!

P.S. He has other notes as well: http://rutherglen.science.mq.edu.au/wchen/ln.html

Cheers!

Answer 9

thanks for the site

Answer 10

@mathteacher, thanks i will look. but there is something in this computation i am missing.
i have to run but will be back. thanks

Answer 11

\( (z+i)^2 = z^2 + 2iz - 1 \) hence
\[ \frac{z^2 - 1}{(z+i)^2} = 1 - \frac{2iz}{(z+i)^2} \]
and
\[ \frac{2iz}{(z+i)^2} = (z-i).\frac{2i}{(z+i)^2} + \frac{-2}{(z+i)^2} \]

Now the Laurent expansions of \( 1/(z+i)^2 \) are easy to evaluate. Find them, multiply by the appropriate constants, shift them down given the powers of (z-i), put it all back together.

Answer 12

ok now i think i got it finally. sorry to be so slow.

Answer 13

i think i was looking for it to be somewhat easier and that got me stuck

Answer 14

This one isn't so bad. Complex analysis is fun and very powerful; hopefully you'll finish your course with that sense.