Evaluate $\int_{|z|=1} \frac{1}{z^2 -\frac{3}{2}z + 1} dz$ Evaluate :
$$\int_{|z|=1} \frac{1}{z^2 -\frac{3}{2}z + 1} dz$$
Using residue method :
$$z=\frac{3}{4} \pm i \frac{\sqrt{7}}{4}$$
The problem is the two roots on the boundary $|z|=1$
 A: 
As stated in the comments, the integral $\oint_{|z|=1}\frac{1}{z^2-\frac32 z+1}\,dz$ is not defined.  

However, we can define the Cauchy Principal Value of the integral as the limit  
$$\begin{align}
\text{PV}\left(\oint_{|z|=1}\frac{1}{z^2-\frac32 z+1}\,dz\right)&=\lim_{\epsilon \to 0}\int_{C_\epsilon} \frac{1}{z^2-\frac32 z+1}\,dz\tag 1
\end{align}$$
where $C_\epsilon$ is the contour $|z|=1$, $\arg(z)\notin (\pm \arctan(\sqrt{7}/3)-\epsilon,\pm \arctan(\sqrt{7}/3)+\epsilon)$. That is to say that $C_\epsilon$ excludes the poles of the integrand.

We can evaluate the limit in $(1)$ by using Cauchy's Integral Theorem.  We close the contour $C_\epsilon$ with contours that are circular arcs around the poles such that neither pole is enclosed. 
As $\epsilon\to 0^+$ of the integration of the arc around the pole at $z=\frac34+i\frac{\sqrt 7}{4}$ can be easily evaluated using the parametric description $z=\frac34+i\frac{\sqrt 7}{4}+2\sin(\epsilon/2)e^{i\phi}$, $\pi/2 +\epsilon<\phi<3\pi/2-\epsilon$.  Proceeding we have 
$$\lim_{\epsilon\to 0^+}\int_{\pi/2 +\epsilon}^{3\pi/2-\epsilon}\frac{i}{2i\frac{\sqrt 7}{4}+2\sin(\epsilon/2)e^{i\phi}}\,d\phi=\frac{2\pi}{\sqrt 7}$$
Similarly, for the integration of the arc around the pole at $z=\frac34-i\frac{\sqrt 7}{4}$ we find
$$\lim_{\epsilon\to 0^+}\int_{\pi/2 +\epsilon}^{3\pi/2-\epsilon}\frac{i}{-2i\frac{\sqrt 7}{4}+2\sin(\epsilon/2)e^{i\phi}}\,d\phi=-\frac{2\pi}{\sqrt 7}$$
Obviously, since the sum of the residues is zero, then we find

$$\text{PV}\left(\oint_{|z|=1}\frac{1}{z^2-\frac32 z+1}\,dz\right)=0 \tag 2$$


Another interpretation is to take the average of $\oint_{|z|=1+\epsilon}\frac{1}{z^2-\frac32 z+1}\,dz$ and $\oint_{|z|=1-\epsilon}\frac{1}{z^2-\frac32 z+1}\,dz$ and then take the limit as $\epsilon\to 0^+$. 
Since the contour $|z|=1-\epsilon$ encloses neither pole, Cauchy's Integral Theorem guarantees that 
$$\oint_{|z|=1-\epsilon}\frac{1}{z^2-\frac32 z+1}\,dz=0 \tag 3$$
And since the contour $|z|=1+\epsilon$ encloses both poles we find from the reside theorem that 
$$\begin{align}
\oint_{|z|=1\epsilon}\frac{1}{z^2-\frac32 z+1}\,dz&=2\pi i \text{Res}\left(\frac{1}{z^2-\frac32 z+1}, z=\frac34\pm i \frac{\sqrt 7}{4}\right)\\\\
&=2\pi i \left(\frac{1}{2i\frac{\sqrt 7}{4}}+\frac{1}{-2i\frac{\sqrt 7}{4}}\right)\\\\
&=0\tag 4
\end{align}$$
The average of $(3)$ and $(4)$ is also $0$ and we find 

$$\text{PV}\left(\oint_{|z|=1}\frac{1}{z^2-\frac32 z+1}\,dz\right)=0 \tag 5$$

And we see that $(5)$ agrees with $(2)$, as expected!
