Question

Laurent series

Answer 1

\[f(z)= \frac{ 1 }{ z(z-i)^2 }\]

Answer 2

when z=i

Answer 3

\[t=z-i; f(z)=\frac{ 1 }{ (t+i )t^2}\]

Answer 4

t=0. And no idea after this! 1/t+i is analytic, but i don't see any power series associated with it

Answer 5

Is this valid?\[\frac{ 1 }{ t^2 }(\frac{ 1 }{ i }(\frac{ 1 }{ i+(t/i) }))\]

Answer 6

\[\frac{ 1 }{ t^2i }\sum_{n=0}^{\infty}(\frac{ -t }{ i })^n\]

Answer 7

@zepdrix @AlexandervonHumboldt2

Answer 8

$$f(z)=\frac1{z(z−i)^2}$$
use a partial fraction decomposition: $$\frac1{z(z-i)^2}=\frac{A}z+\frac{B}{z-i}+\frac{C}{(z-i)^2}\\\begin{align*}1&=A(z-i)^2+Bz(z-i)+Cz\\&=A(z^2-2iz-1)+B(z^2-iz)+Cz\\&=(A+B)z^2+(-2Ai-Bi+C)z-A\end{align*}\\\implies A=-1, B=1, C=-i$$

Answer 9

so $$-\frac1z=-\frac1i\cdot\frac1{1+(z-i)/i}=-\frac1i\cdot\sum_{n=0}^\infty\left(-\frac{z-i}i\right)^n=\sum_{n=0}^\infty i^{n+1} (z-i)^n$$

Answer 10

and obviously $$\frac1{z-i}=(z-i)^{-1}\\\frac{-i}{(z-i)^2}=-i(z-i)^{-2}$$ so the Laurent expansion is $$\frac1{z(z-i)^2}=-\frac{i}{(z-i)^2}+\frac1{z-i}+\sum_{n=0}^\infty i^{n+1} (z-i)^n$$

Answer 11

note, we could've also just looked at $$\frac1z=\sum_{n=0}^\infty i^{n-1}(z-i)^n$$ and realized that $$\frac1{z(z-i)^2}=\frac1{(z-i)^2}\cdot\sum_{n=0}^\infty i^{n-1}(z-i)^n=\sum_{n=0}^\infty i^{n-1}(z-i)^{n-2}=\sum_{n=-2}^\infty i^{n+1}(z-i)^n$$

Answer 12

also, you could represent \(\frac1{t+i}\) using the same trick I did above: $$\frac1{t+i}=\frac1i\cdot\frac1{1-(-t/i)}=\frac1i\sum_{n=0}^\infty (-t/i)^n=-\sum_{n=0}^\infty i^{n+1} t^n$$

Answer 13

shouldn't \[\frac{ 1 }{ i }\sum_{}^{}(-t/i)^n =\sum_{}^{} \frac{ -t^n }{ i^{n+1} }\]

Answer 14

yes, and \(-1/i=i\)