Question

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 ______.

• A. \(6\pi j\)
• B. \(3\pi j\)
• C. \(12\pi j\)
• D. \(\pi j\)

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: \(\oint_{\gamma}\dfrac{f(z)}{z-z_0}dz=2\pi j\,f(z_0)\).

Answer 1

The Correct Option is A

Step 1: Identify the form of the integral.
The integral is
\[ \oint_{\gamma}\frac{f(z)}{z-1}\,dz \] where \(f(z)=z^2+z+1\) is analytic everywhere, since it is a polynomial with no singularities anywhere in the complex plane. The only place the integrand \(\dfrac{f(z)}{z-1}\) can fail to be analytic is at \(z=1\), which is a simple pole because the denominator vanishes there to first order.

Step 2: Recall Cauchy's Integral Formula.
Cauchy's Integral Formula states that if \(f(z)\) is analytic inside and on a simple closed contour \(\gamma\), and \(z_0\) is any point strictly inside \(\gamma\), then
\[ \oint_{\gamma}\frac{f(z)}{z-z_0}\,dz=2\pi j\,f(z_0) \] This formula lets us evaluate the integral just by plugging \(z_0\) into \(f\), without carrying out any actual integration, as long as \(z_0\) is the only singularity of the integrand inside the contour.

Step 3: Match the given integral to the formula.
Here \(z_0=1\), which is given to lie inside \(\gamma\), and \(f(z)=z^2+z+1\) is analytic everywhere, including inside and on \(\gamma\). So the formula applies directly with \(z_0=1\).

Step 4: Evaluate \(f\) at the pole.
\[ f(1)=1^2+1+1=3 \]

Step 5: Substitute into Cauchy's formula.
\[ \oint_{\gamma}\frac{f(z)}{z-1}\,dz=2\pi j\,f(1)=2\pi j\times3=6\pi j \]

Step 6: Rule out the other options.

(B) \(3\pi j\): This would follow only if one forgot to double \(f(1)\), using \(\pi j\times f(1)\) instead of \(2\pi j\times f(1)\). Incorrect.

(C) \(12\pi j\): This would follow from mistakenly using \(4\pi j\,f(1)\) instead of \(2\pi j\,f(1)\). Incorrect.

(D) \(\pi j\): This is far too small and does not match \(2\pi j\,f(1)\) for the correct value of \(f(1)\). Incorrect.

Final Answer:
\[ \boxed{6\pi j} \]