The value of the integral

$\oint \left(\frac{6\mathrm{z}}{2{\mathrm{z}}^4-3{\mathrm{z}}^3+7{\mathrm{z}}^2-3\mathrm{z}+5}\right)\mathrm{d}\mathrm{z}$

evaluated over a counter-clockwise circular contour in the complex plane enclosing only the pole 𝑧 = 𝑖, where 𝑖 is the imaginary unit, is

The correct option/answer is: (−1 + 𝑖) π

To find the value of the given complex integral, we can apply the Cauchy Residue Theorem.

The residue theorem states that if a function $f\left(z\right)$ is analytic inside and on a simple closed contour $C$ except at a finite number of isolated singularities (poles) inside $C$, then the integral is given by:

${\oint }_{C}f\left(z\right)dz=2\pi i\sum _{k=1}^n{\text{Res}}_{z=z_k}f\left(z\right)$

In this problem, the contour is a counter-clockwise circle enclosing only the pole $z=i$. Therefore, the integral evaluates to:

$I=2\pi i\cdot {\text{Res}}_{z=i}f\left(z\right)$

where the integrand is:

$f\left(z\right)=\frac{6z}{2z^4-3z^3+7z^2-3z+5}$

Let us denote the denominator as $P\left(z\right)=2z^4-3z^3+7z^2-3z+5$.
First, let's verify that $z=i$ is a root of the denominator $P\left(z\right)$:

$P\left(i\right)=2(i{)}^4-3(i{)}^3+7(i{)}^2-3(i)+5$

Recall that $i^2=-1$, $i^3=-i$, and $i^4=1$. Substituting these values in, we get:

$P\left(i\right)=2\left(1\right)-3(-i)+7(-1)-3i+5$

$P\left(i\right)=2+3i-7-3i+5=0$

Thus, $z=i$ is indeed a zero of the denominator.
To determine if it is a simple pole (order 1), we find the derivative of $P\left(z\right)$:

$P^{\prime }\left(z\right)=8z^3-9z^2+14z-3$

Evaluating this derivative at $z=i$:

$P^{\prime }\left(i\right)=8(i{)}^3-9(i{)}^2+14\left(i\right)-3$

$P^{\prime }\left(i\right)=8(-i)-9(-1)+14i-3$

$P^{\prime }\left(i\right)=-8i+9+14i-3=6+6i$

Since $P^{\prime }\left(i\right)\ne 0$, $z=i$ is a simple pole of $f\left(z\right)$.

For a function of the form $f\left(z\right)=\frac{g\left(z\right)}{h\left(z\right)}$ with a simple pole at $z=z_0$, the residue is computed as:

${\text{Res}}_{z=z_0}f\left(z\right)=\frac{g\left(z_0\right)}{h^{\prime }\left(z_0\right)}$

Substituting our values for the numerator $g\left(z\right)=6z$ and the derivative of the denominator $P^{\prime }\left(z\right)$ at $z=i$:

${\text{Res}}_{z=i}f\left(z\right)=\frac{6\left(i\right)}{6+6i}=\frac{i}{1+i}$

We can simplify this complex fraction by multiplying the numerator and denominator by the complex conjugate of the denominator, $1-i$:

${\text{Res}}_{z=i}f\left(z\right)=\frac{i(1-i)}{(1+i)(1-i)}=\frac{i-i^2}{1-i^2}=\frac{i-(-1)}{1-(-1)}=\frac{1+i}{2}$

Now, calculate the final value of the contour integral:

$I=2\pi i\cdot \left(\frac{1+i}{2}\right)$

$I=\pi i(1+i)=\pi (i+i^2)=\pi (i-1)=(-1+i)\pi$