How to come from $\frac{z^{2}}{z^{2}+1}$ to $\frac{z/2}{z+i}+\frac{z/2}{z-i}$

Question

As title says, how to come from $\frac{z^{2}}{z^{2}+1}$ to $\frac{z/2}{z+i}+\frac{z/2}{z-i}$?

Here is what I did: $\frac{z^{2}}{z^{2}+1}=1-\frac{1}{z^{2}+1}=1-\frac{1}{(z-i)(z+i)}$

Answer 1

$$\frac{z^2}{z^2+1} = z\cdot \frac{z}{z^2+1}$$

Now, use partial fractions to get $$\frac z{z^2+1} = \frac{\frac12}{z+i} + \frac{\frac12}{z-i}$$ and plug that into the first equation.

Answer 2

Firstly $$\frac{z^{2}}{z^{2}-1} = \frac{Az}{z+i}+\frac{Bz}{z-i}$$ now cross multiply by $(z+i)(z-i)$ are then take specific values of $z= \pm i$ to find specific values for $A, B$.

Edit I trust is is clear why we split the denominators in that way. I won't go into details as I'll assume that step is trivial. To see why I chose numerators above as simply $Az$ and $Bz$ as noted by @5xum below; don't do that. Take $$\frac{z^{2}}{z^{2}+1} = \frac{az+b}{z-i}+\frac{cz+d}{z+i}$$

Now cross multiply as I mentioned above giving, $$z^{2}=(z+i)(az+b)+(z-i)(cz+d)$$ Choose specific values for $z$. Namely, the zeroes of the denominators. Choose $z=i$ \begin{align} -1 &= 2i(ai+b)+(i-i)(cz+d) \\ \implies -1 &= 2i(ai+b) \\ \implies \frac{i}{2} &= ai+b \end{align} This relation shows that $b=0$ since the left hand side is purely imaginary. Thus $a=\frac{1}{2}$. Repeating gives similar expression for $d$ and your result follows.

Answer 3

Make the ansatz $$\frac{z^{2}}{z^{2}+1} =\frac{a}{z+i}+\frac{b}{z-i} =\frac{a(z-i)+b(z+i)}{(z-i)(z+i)} =\frac{(a+b)z-(a-b)i}{(z-i)(z+i)}$$ Now solve for $a-b=0$ and $a+b=z$. The first gives $a=b,\,$ substitute into the second gives $2a=z,\,$ i.e. $a=b=\frac{z}{2}.$