Find the residue at each of the isolated singularities of the following function on $\mathbb{C}$?
$1) \frac{z}{z^2+3z +3}$
$ii) \frac{1}{(z^3 +1)(z+1)^2}$
My attempt : for $1) z^2 +3z + 3= z^2 + 2z+z + 3$ where $z = \frac{-3 +\sqrt 3^2 -12}{2}$ I know that here $z$ is a simple pole
For II) $f(z)=\frac{1}{(z^3 +1)(z+1)^2}$ here $z=-1$ is pole of order $2 $
Any hints/solution will be appreciated
thanks u
For your second problem, $f(z)= \frac{1}{(z^3 +1)(z+1)^2}$, you have $z=-1$ as a pole of order 3, and then 2 simple poles (can you see what they are?).
To calculate your pole of order 3, use the residue formula. You'll have $$\frac{1}{(3-1)!} \frac{d^2}{dz^2} (z+1)^3 f(z)$$. Then after differentiating, take the limit as it approaches the pole, $-1$. Can you take it from here?