<s>[INST] Find the residue of $ \bigl(\frac{z}{2z+1}\bigr)^3$ Find the residue of : $\displaystyle \left(\frac{z}{2z+1}\right)^3$
In this case we have $-\frac{1}{2}$ as a pole. Now I want to use the following formula:
$\displaystyle \frac{\phi^{k-1}(z_{0})}{(k-1)!}$
Now would I let $k = 3$ in this case then? 
After that I am a little unsure what to do. The answer is $\displaystyle-\frac{3}{16}$ .
 [/INST] Here is a way to not use that formula.  Note that $$\dfrac{z}{2z+1}=\dfrac{1}{2}\cdot\frac{z}{z+\frac12}=\frac{1}{2}\left(1-\frac{1}{2}\cdot\frac{1}{z+\frac{1}{2}}\right)=\frac12-\frac14\cdot\frac{1}{z+\frac12}.$$
You can use $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ with $a=\dfrac12$ and $b=-\dfrac14\cdot\dfrac{1}{z+\frac12}$ to see that $\left(\dfrac{z}{2z+1}\right)^3 = -\dfrac{3}{16}\cdot\dfrac{1}{z+\frac12} + $ other stuff.
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