Examine the behaviour of the integral

Question

I have a hint to write the ratio $ \frac{{z}^{4}\mathrm{{+}}{z}^{2}\mathrm{{+}}{1}}{{z}{\mathrm{(}}{z}^{2}\mathrm{{+}}{1}{\mathrm{)}}} $

Like this : $ \frac{{z}^{4}\mathrm{{+}}{z}^{2}\mathrm{{+}}{1}}{{z}{\mathrm{(}}{z}^{2}\mathrm{{+}}{1}{\mathrm{)}}}\mathrm{{=}}{z}\mathrm{{+}}\frac{1}{z}\mathrm{{+}}\frac{1}{2}{\mathrm{(}}\frac{1}{{z}\mathrm{{+}}{i}}\mathrm{{+}}\frac{1}{{z}\mathrm{{-}}{i}}{\mathrm{)}} $

Then how can I examine the behaviour of the integral $ \mathop{\int}\limits_{\left|{{z}\mathrm{{-}}{i}}\right|\mathrm{{=}}{R}}{}\frac{{z}^{4}\mathrm{{+}}{z}^{2}\mathrm{{+}}{1}}{{z}{\mathrm{(}}{z}^{2}\mathrm{{+}}{1}{\mathrm{)}}} $

As a function of $ {R}\mathrm{{>}}{0} $

Answer 1

Since we can write

$$\frac{z^4+z^2+1}{z(z^2+1)}=z+\frac1z+\frac12\left(\frac{1}{z+i}+\frac{1}{z-i}\right)$$

we have

$$\begin{align} \oint_{|z-i|=R}\frac{z^4+z^2+1}{z(z^2+1)}&=\oint_{|z-i|=R}z\,dz \tag 1\\\\ &+\oint_{|z-i|=R}\frac1z\,dz\\\\ &+\frac12\oint_{|z-i|=R}\frac{1}{z+i}\,dz\\\\ &+\frac12\oint_{|z-i|=R}\frac{1}{z-i}\,dz \end{align}$$

We can evaluate each of the integrals in $(1)$ by using Cauchy's Integral Formula.

The first integral on the right-hand side of $(1)$ is $0$ since $z$ is holomorphic.

The second integral on the right-hand side of $(1)$ is $0$ for $R<1$, and $2\pi i$ for $R>1$.

The third integral on the right-hand side of $(1)$ is $0$ for $R<2$, and (with the factor of $1/2$) $\pi i$ for $R>2$.

The last integral (with the factor of $1/2$) is $\pi i$.

Can you put all of this together now?