Calculation of Residue. Let $c=\cos \dfrac{\pi}{5}, f(z)=\dfrac{z^2-2cz+1}{z^4-z^3+z^2-z+1}$.
$e^{\frac{3\pi}{5}i}$ is one of the pole of $f$. (This is because $f$ can be written as $f(z)=\frac{(z+1)(z^2-2cz+1)}{z^5+1}$.)
Then, calculate the Residue of $f$ at $e^{\frac{3\pi}{5}i}=:a$.

I calculated using the formula of Residue, but the calculation is complicated and I don't know how I should proceed.

\begin{align}
\mathrm{Res}(f,a)
&=\displaystyle\lim_{z\to a} (z-a)f(z)\\
&=\lim_{z\to a} \dfrac{(z-a)(z^2-2cz+1)}{z^4-z^3+z^2-z+1}\\
&=\lim_{z\to a}\dfrac{z^2-2cz+1+(z-a)(2z-2c)}{4z^3-3z^2+2z-1}\\
&=\dfrac{a^2-2ca+1}{4a^3-3a^2+2a-1}.
\end{align}
I have to simplify this, but I don't know how I can do.
I think I have to use some technical method. Thanks for any idea.

We can considerably simplify the residue calculation when expanding numerator and denominator with $z+1$. This way we can effectively get rid of the denominator. We consider
\begin{align*}
f(z)=\frac{(z+1)\left(z^2-2cz+1\right)}{z^5+1}
\end{align*}
and obtain

\begin{align*}
\mathrm{Res}&(f,a)
=\displaystyle\lim_{z\to a} (z-a)f(z)\\
&\color{blue}{=\lim_{z\to a}}\color{blue}{\frac{(z-a)(z+1)\left(z^2-2cz+1\right)}{z^5+1}}\\
&=\lim_{z\to a}\frac{\left(z^2+(1-a)z-a\right)\left(z^2-2cz+1\right)}{z^5+1}\\
&=\lim_{z\to a}\frac{\left(2z+(1-a)\right)\left(z^2-2cz+1\right)+\left(z^2+(1-z)z-a\right)(2z-2c)}{5z^4}\tag{1}\\
&=\frac{(a+1)\left(a^2-2ca+1\right)}{5a^4}\tag{2}\\
&=-\frac{1}{5}a(a+1)\left(a^2-2ca+1\right)\tag{3}\\
&=-\frac{1}{5}\left(a^4+(1-2c)a^3+(1-2c)a^2+a\right)\\
&=-\frac{1}{5}\left(a+a^4\right)-\frac{1}{5}\left(1-2c\right)\left(a^2+a^3\right)\\
&\color{blue}{=-\frac{1}{5}\left(a-\frac{1}{a}\right)-\frac{1}{5}(1-2c)\left(a^2-\frac{1}{a^2}\right)}\tag{4}
\end{align*}
Since
\begin{align*}
a-\frac{1}{a}&=\exp\left(3 i\pi/5\right)-\exp\left(-3 i\pi/5\right)=2i\sin(3\pi/5)\\
a^2-\frac{1}{a^2}&=-\exp\left(i\pi/5\right)+\exp\left(-i\pi/5\right)=-2i\sin(\pi/5)\\
\end{align*}
and noting that
\begin{align*}
c=\cos(\pi/5)=\frac{1}{4}\left(1+\sqrt{5}\right)
\end{align*}
it shouldn't be too hard to finish the calculation.

Comment:

*

*In (1) we apply L'Hôpital's rule.


*In (2) we calculate the limit noting the right-hand term of the numerator is zero.


*In (3) we use $a\cdot a^4=-1$.


*In (4) we again use properties of the units: $a\cdot a^4=a^2\cdot a^3=-1$.
