I wish to construct a function $f(z)$ with the properties as
- only singularities of $f(z)$ in the extended complex plane are poles of order of $1$ and $2$ at $z= 1$ and $z=-1$ respectively.
- Also it is given that $f(0)= 0=f(-1/2)$ and residue of $f(z)$ at $z= 1$ and $z=-1$ is equal to $1$ .
Sketch:
Write your function as a rational function. This makes it easy to specify the poles and zeros (when you have finitely many).
The information that $z=1$ is a pole of order 1 and $z=-1$ is a pole of order $2$ implies that your function appears as (where the $\cdots$ indicates that we have not filled in the numerator yet. $$ \frac{\cdots}{(z-1)(z+1)^2}. $$
Since $f(0)=0=f(-1/2)$, then the numerator includes the factors $z$ and $2z+1$. In other words, our function is of the form (where the $\cdots$ indicates that we might have more to fill in). $$ \frac{z(2z+1)\cdots}{(z-1)(z+1)^2}. $$
Since the residue at $z=1$ is $1$, then substituting $z=1$ into $$ \frac{z(2z+1)\cdots}{(z+1)^2} $$ should give you $1$.
Since the residue at $z=-1$ is $1$, and this is a pole of order $2$, then substituting $z=-1$ into $$ \frac{d}{dz}\frac{z(2z+1)\cdots}{(z-1)} $$ should also give you $1$.
(Check these last two formulas on your own.)
Now, all you need to do is to figure out any appropriate $\cdots$ that make the last two equations true. (and don't change the order of the poles).
Responding to comments:
The last two conditions correspond to two equations, so we would expect to need two variables in the $\cdots$ to be able to simultaneously solve them. Therefore, we would look at $$ \frac{z(2z+1)(az+b)}{(z-1)(z+1)^2}. $$
Using this form to guess, the two equations become: $$ \left.\frac{z(2z+1)(az+b)}{(z+1)^2}\right|_{z=1}=\frac{3(a+b)}{4}=1 $$ and $$ \left.\frac{d}{dz}\right|_{z=-1}\frac{z(2z+1)(az+b)}{(z-1)}=\left.\frac{(z-1)((2z+1)(az+b)+2z(az+b)+az(2z+1))-z(2z+1)(az+b)}{(z-1)^2}\right|_{z=-1}=1. $$ Then, solve for $a$ and $b$.