Complex integral of $\int_{\partial D(0,1)}\frac{\mathrm{d}z}{z^2 -3z +1} $

Question

I have to calculate $\int_{\partial D(0,1)}\frac{\mathrm{d}z}{z^2 -3z +1} = \int_0^1 \frac{2 i \pi e^{2 i \pi t}\mathrm{d}t}{(e^{2 i \pi t})^2 -3e^{2 i \pi t} +1}=\int_0^{2\pi} \frac{ i e^{i \theta}\mathrm{d}\theta}{(e^{i \theta})^2 -3e^{i \theta} +1}$ and I found a $tanh^{-1}$ as antiderivative and I found zero as final result. But Wolfram Alpha find $-\frac{2 i \pi}{\sqrt{5}}$ as final result. I really don't understand where my mistake could be. Can you help me?

Answer 1

Let's $r1=\frac{3-\sqrt{5}}{2}$ and $r2=\frac{3+\sqrt{5}}{2}$ be the roots of $z^2 -3z +1$, then we have \begin{align*} \int_{\partial D(0,1)}\frac{\mathrm{d}z}{z^2 -3z +1} = -\frac{1}{\sqrt{5}} \left(\int_{\partial D(0,1)}\frac{1}{z-r1}\mathrm{d}z - \int_{\partial D(0,1)}\frac{1}{z-r2}\mathrm{d}z \right) \end{align*}

But $r2=\frac{3+\sqrt{5}}{2}>1$ so $r2 \notin D(0,1)$ then $\int_{\partial D(0,1)}\frac{1}{z-r2}\mathrm{d}z=0$

Let's calculate $\int_{\partial D(0,1)}\frac{1}{z-r1}\mathrm{d}z$ : we set $g(z)=1$, from integral Cauchy formula, we get

\begin{align*} &g(r1) =\frac{1}{2 \pi i} \int_{\partial D(0,1)}\frac{g(z)}{z-r1}\mathrm{d}z\\ \iff & 1 = \frac{1}{2 \pi i} \int_{\partial D(0,1)}\frac{g(z)}{z-r1}\mathrm{d}z\\ \iff &\int_{\partial D(0,1)}\frac{g(z)}{z-r1}\mathrm{d}z = 2 \pi i\\ \iff &\int_{\partial D(0,1)}\frac{1}{z-r1}\mathrm{d}z = 2 \pi i\\ \end{align*}

We deduce

$$ \int_{\partial D(0,1)}\frac{\mathrm{d}z}{z^2 -3z +1} = \frac{-2 \pi i}{\sqrt{5}} $$