# Partial fraction decomposition of the given equation

How do I do the partial fraction decomposition for $$\frac{z}{z^2+1}$$

This is in terms of complex analysis. I know how to do this in terms of $$\frac{1}{z^2+1}$$, but not sure what to do about it when the numerator is z instead of 1.

FYI, I know that $$z^2+1 = (z+i)(z-i)$$, so the above can be re-expressed as $$\frac{z}{z^2+1}= \frac{z}{(z+i)(z-i)}$$

You know: $$\frac{z}{z^{2}+1}=\frac{z}{(z+i)(z-i)}=\frac{a}{z+i}+\frac{b}{z-i}$$

so we multiply both sides by $$z^2+1$$ and get: $$a(z-i)+b(z+i)=z$$

Set $$z=i$$ to solve for $$b$$ and $$z=-i$$ to solve for a and you get:

$$\frac{1}{2}\left(\frac{1}{z+i}+\frac{1}{z-i}\right)$$

($$a=b=\frac{1}{2}$$)

You know that $$\frac z{z^2+1}$$ can be written as$$\frac a{z+i}+\frac b{z-i}=\frac{(a+b)z+i(-a+b)}{z^2+1}.$$So, solve the system$$\left\{\begin{array}{l}a+b=1\\-a+b=0.\end{array}\right.$$

• If you are restricted to real numbers, then there is NO simpler "partial fractions" – user247327 Oct 14 '19 at 20:42