How to find the value of the integral?

Question

Find the value of the integral $$\int_C \frac {3z^5 - 10z^3} {z^6-5z^4+10}\ dz$$

where $C = \left \{z \in \Bbb C\ :\ |z| < 2 \right \}.$

I know that $$\int_{\gamma} f(z)\ dz = \int_{a}^{b} f \left (\gamma (t) \right ) \gamma'(t)\ dt$$

where $\gamma : [a,b] \longrightarrow \mathbb C$ be a piecewise continuous path (called a contour). By using this formula the given integral takes a weird form which I unable to simplify.

Would anybody please help me in this regard? Thank you very much.

Answer 1

I believe there are $4$ poles inside the region. Using Rouche's theorem, the denominator has the same number of zeros in $C$ as $-5z^4$: $|z^6+10|\le|5z^4|=80$.

The residue at each pole is $p(z_k)/q'(z_k)=1/2$.

Thus by the residue theorem we get $2\pi i(4\cdot 1/2)=4\pi i$.

Answer 2

First solution: Computation using the definition, substitution $z=2e^{it}$: $$ \begin{aligned} J &:= \int_C \frac {3z^5 - 10z^3} {z^6-5z^4+10}\; dz \\ &=\int_0^{2\pi} \frac {3\cdot 2^5 e^{5it} - 10\cdot 2^3e^{3it}} {2^6\cdot e^{6it}-5\cdot 2^4 e^{4it}+10} \;2i\; e^{it}\; dt \\ &= \frac 12 \int_0^{2\pi} \Big[\ \ln(\ 2^6\cdot e^{6it}-5\cdot 2^4 e^{4it}+10\ )\ \Big]' \; dt \\ &= \frac 12 \Big[\ \ln(\ 2^6\cdot e^{6it}-5\cdot 2^4 e^{4it}+10\ )\ \Big]_0^{2\pi} \\ &=\dots \end{aligned} $$ Now let us take a look at a plot of the function under the logarithm:

It starts in $32-80+10=-38$, and maps the interval $[0,\pi/2]$ into

then it closes the contour as in the picture on $[0,\pi]$, and then the corresponding even function $z^6-5z^4+10$ makes a new tour on it on $[\pi, 2\pi]$ using the parametrization $z=2e^{it}$. Taking logarithm, and computing the difference for the values in $t=0$ and $t=2\pi$ reveals the corresponding monodromy of the logarithm, we have to consier four times $2\pi i$. So the result is $\frac 12\cdot 8\pi i=4\pi i$.

Second solution: Residue computation.

First of all, let us locate all roots of the polynomial in the denominator, computer aided information:

sage: g = z^6 - 5*z^4 + 10
sage: roots = g.roots(ring=QQbar, multiplicities=0)
sage: roots.sort()
sage: roots
[-2.123182227548205?,
 -1.325005689057003?,
 -1.124074217927793?*I,
 1.124074217927793?*I,
 1.325005689057003?,
 2.123182227548205?]

Let $a$ be a root under the above numbers inside the circle $C$. Then it is a pole of order one for the function under the integral, and the residue is $$ \lim_{z\to a} (z-a)\cdot \frac {3z^5 - 10z^3} {z^6-5z^4+10} = \lim_{z\to a} \frac {3z^5 - 10z^3} {(z^6-5z^4+10)/(z-a)} = \frac {3a^5 - 10a^3} {(z^6-5z^4+10)'_{\text{in }z=a} } =\frac 12\ . $$ There are four roots of the denominator inside the circle $C$, so by the residue theorem the value of the integral is $$ 2\pi i\left( \frac 12+ \frac 12+ \frac 12+ \frac 12 \right) =4\pi i\ . $$

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P.S. The answer was started last year... now i could complete and submit...