Question

how about a nice laurant expansion
\[\frac{z}{z^2+1}\text { at } z_0=i\]

Answer 1

Is this pre-calculus or something higher?

Answer 2

i have a actually sort of computed this beast, but i am wondering if there is some snappier way to do it, because calculation was a total drag and probably wrong at least by a sign

Answer 3

Oh nvm

Answer 4

a bit higher, and quite a bit more annoying.

Answer 5

started with
\[\frac{1}{z-i}\frac{z}{z+i}=\frac{1}{z-i}(1-\frac{i}{z+i})=\frac{1}{z-i}(1-\frac{i}{2i+(z-i)})\] then second part is
\[1-\frac{\frac{1}{2}}{1-\frac{(z-i)}{2i}}\]

Answer 6

maybe there is a nice snap method, maybe...

Answer 7

I'm thinkin
\[ u = z - i \rightarrow z = u+ i \rightarrow \frac{u+i}{(u+i)^2+1} = \frac{u+i}{u^2+2iu}\]
\[= \frac{1}{u+2i} + \frac{1}{u}\frac{i}{u+2i} = \left(1+\frac{i}{u}\right)\frac{1}{u+2i}\]
then, if mod(u) < 2i, then we get

\[\left(1+\frac{i}{u}\right) \frac{1}{2i} \frac{1}{1+\frac{u}{2i}}=\left(1+\frac{i}{u}\right)\sum_{n=0}^\infty \frac{u^n}{(2i)^{n+1}}\]

and if mod(u) > 2i, we get
\[\left(1+\frac{i}{u}\right)\frac{1}{u} \sum_{n=0}^\infty \frac{(2i)^n}{u^{n}}=\left(1+\frac{i}{u}\right)\sum_{n=0}^\infty \frac{(2i)^n}{u^{n+1}} \]
They can be simplified a little further from there, but I'm lazy.

Answer 8

Oh, and then of course make the substitution u = z-i...

Answer 9

Be nice if you could explain exactly what you did in detail

Answer 10

Just post the explanation. I'll read it tomorrow. I'm tired and got horse cough

Answer 11

Oops! I made a mistake anyway :) put a (-1)^n in both of those summations... but here's the actual explanation....

I want to create something of the form
\[\frac{1}{1-x} , |x|<1\]
so I can immediately turn that into the series
\[\sum_{n=0}^\infty \space x^n \]
As stated above, I substituted z for u+i and did some manipulation to get
\[ \left(1+\frac{i}{u}\right)\frac{1}{u + 2i} \]
NOW
if |u| < 2, I can get
\[ \left( 1+ \frac{i}{u}\right)\cdot \frac{1}{2i} \cdot \frac{1}{1 - (\frac{-u}{2i})} =\left( 1+ \frac{i}{u}\right)\cdot \frac{1}{2i} \cdot \sum_{n=0}^\infty (-1)^n\left(\frac{u}{2i}\right)^n \]
if, on the other hand, |u| > 2, then we have to use
\[\left(1+\frac{i}{u}\right)\cdot \frac{1}{u}\cdot \frac{1}{1-(\frac{-2i}{u})} =\left(1+\frac{i}{u}\right)\cdot \frac{1}{u}\cdot \sum_{n=0}^\infty(-1)^n\left(\frac{2i}{u}\right)^n\]
It takes a bit of algebra, but those two simplify to the following: (z)/(z^2+1)
\[ \frac{z}{z^2+i} = \frac{1}{2(z-i)} - \frac{i}{4}\sum_{n=0}^\infty(-1)^n\left(\frac{z-i}{2i}\right)^n\]
for |z-i| < 2
and
\[\frac{z}{z^2+1} = \frac{1}{z-i} + \frac{i}{4}\sum_{n=2}^\infty(-1)^n\left(\frac{2i}{z-i}\right)^n \]

Answer 12

The first expansion is therefore valid when |z-i| < 2 and the second is valid when |z-i| > 2. Neither are valid when |z-i| = 2, which of course we could recognize ahead of time because our function is singular at i and -i, and the distance between those two points is 2. Two regions with different Laurent expansions separated by a singularity .... ahhh I miss complex analysis.