Consider a function \(f(z)=z^2+z+1\), where \(z\in\mathbb{C}\) is a complex variable. A simple closed contour \(\gamma\) in the \(z\)-plane encloses the point \(z=1+0j\). The value of the integral \[\oint_{\gamma}\frac{f(z)}{z-1}\,dz\] is ______.
Show Hint
Since \(f(z)\) is a polynomial (analytic everywhere) and the only singularity of \(\dfrac{f(z)}{z-1}\) inside \(\gamma\) is the simple pole at \(z=1\), apply Cauchy's Integral Formula: \(\oint_{\gamma}\dfrac{f(z)}{z-z_0}dz=2\pi j\,f(z_0)\).
Step 1: Locate the singularity inside the contour.
The integrand $\dfrac{f(z)}{z-1}$ with $f(z)=z^2+z+1$ has a single simple pole at $z=1$, since $f(z)$ is a polynomial and never blows up on its own. As $\gamma$ is stated to enclose $z=1$, this is the only pole we need inside the contour.
Step 2: Find the residue at that pole.
For a simple pole at $z=z_0$, the residue of $\dfrac{f(z)}{z-z_0}$ is just $f(z_0)$, because the residue picks out the coefficient of $\dfrac{1}{z-z_0}$ in the Laurent expansion, and that coefficient is exactly $f(z_0)$ when $f$ is analytic at $z_0$. So
\[ \text{Res}_{z=1}\frac{f(z)}{z-1}=f(1)=1+1+1=3 \]
Step 3: Apply the Residue Theorem.
The Residue Theorem says a closed contour integral equals $2\pi j$ times the sum of the residues of the poles enclosed:
\[ \oint_{\gamma}\frac{f(z)}{z-1}\,dz=2\pi j\sum \text{Res}=2\pi j(3) \]
Step 4: Work out the number.
\[ 2\pi j\times3=6\pi j \]
Step 5: Sanity check against the wrong options.
Getting $3\pi j$ or $12\pi j$ would mean using a multiplier of $\pi j$ or $4\pi j$ instead of $2\pi j$, and $\pi j$ alone ignores the residue value $3$ altogether. None of those match the residue-theorem computation done here.
\[ \boxed{6\pi j} \]
