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:

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



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$$\left\{\begin{array}{l}a+b=1\\-a+b=0.\end{array}\right.$$

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

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