one more from the complex plane. expand
$$\frac{z^2-1}{(z^2+1)^2}$$ about
$$z_0=i$$

Question

one more from the complex plane. expand 
$$\frac{z^2-1}{(z^2+1)^2}$$ about 
$$z_0=i$$

anonymous · Answer

i have the answer, just not sure of the algebra strategy.  partial fractions?

jamesj · Answer

Like before, pull out the 1/(z-i) factor.  Then do some long division to simplify the remaining expression.

anonymous · Answer

as in 
$$\frac{1}{z-i}\frac{z^2-1}{(z-i)(z+i)^2}$$?

jamesj · Answer

Pull out (z-i)^2.

anonymous · Answer

$$\frac{1}{(z-i)^2}\frac{z^2-1}{(z+i)^2}$$

jamesj · Answer

Yes.  Now, get rid of the z^2 in the numerator and you'll have a cz + d expression for some complex numbers c, d.

anonymous · Answer

ok thanks again. looks like this is just going to be some annoying algebra

anonymous · Answer

i'd rather compute the limit of a reimann sum

jamesj · Answer

lol.  You've seen that?  This strobe guy is real retrice

jamesj · Answer

restrice?  no.  a-s-s.

anonymous · Answer

yeah i got a chuckle from that one

anonymous · Answer

there must be something about math (maybe the fact that one is usually right or wrong) that turns people into opinionated retrices pieces

pokemon23 · Answer

chess time

anonymous · Answer

hmmm this is not working the way i thought it would.  maybe i made a mistake

anonymous · Answer

e4

pokemon23 · Answer

e5

anonymous · Answer

n f3

pokemon23 · Answer

hmmm