Where is $f(z)=\frac{2z+1}{z(z^2+1)}$ analytic?

Question

I have to find the points where $f(z)=\dfrac{2z+1}{z(z^2+1)}$ is analytic.

My idea is, denoting $z=x+\mathrm iy$, we find the functions $u(x,y)$ and $v(x,y)$ first by multiplying the conjugate of the denominator: $$\frac{2z+1}{z(z^2+1)}\cdot\frac{\overline{z^3+z}}{\overline{z^3+z}},$$ then using the Cauchy-Riemann equations to find the points. But, just finding $u(\cdot,\cdot)$ and $v(\cdot,\cdot)$ is an arduous task due to the resulting expansion.

Is this correct, or is there a simpler method (preferably similar) to solve this?

Answer 1

$f'(z)=\dfrac{z(z^2+1)(2)-(2z+1)[(z^2+1)+2z^2]}{z^2(z^2+1)^2}$ exists precisely at the points where the denominator is not zero, namely $z\ne 0,i,-i$. Thus that's where $f(z)$ is analytic.

Answer 2

$f(z)$ is not analytic at its poles.

By looking at the denominator, you can see that it has poles at $z=0,i,$ and $-i$.