Question

If \(C\) is a simple closed curve not passing through the points \(-1\), \(0\) and \(1\) in the complex plane, then the set of all possible value(s) of the integral \(\int_C \dfrac{dz}{z(1-z^2)}\) is ____.

• A. \(\{0,\ \pm i\pi,\ \pm 2i\pi\}\)
• B. \(\{0,\ \pm i\pi,\ 2i\pi\}\)
• C. \(\{0,\ \pm i\pi\}\)
• D. \(\{0\}\)

Hint:

Find the residues at \(0,1,-1\) and note a simple closed curve can enclose any subset of these three points.

Answer 1

The Correct Option is A

Step 1: Factor the denominator: \(z(1-z^2)=z(1-z)(1+z)\). The integrand \(f(z)=\dfrac{1}{z(1-z)(1+z)}\) has simple poles at \(z=0,1,-1\).

Step 2: Compute the residues. At \(z=0\): \(\operatorname{Res}_{z=0}f=\dfrac{1}{(1-0)(1+0)}=1\).

Step 3: At \(z=1\): write \(f(z)=\dfrac{-1}{z(z-1)(1+z)}\), so \(\operatorname{Res}_{z=1}f=\dfrac{-1}{z(1+z)}\Big|_{z=1}=\dfrac{-1}{2}\).

Step 4: At \(z=-1\): \(\operatorname{Res}_{z=-1}f=\dfrac{1}{z(1-z)}\Big|_{z=-1}=\dfrac{1}{(-1)(2)}=\dfrac{-1}{2}\).

Step 5: A simple closed curve \(C\) has a simply connected interior (Jordan curve theorem), so its interior can contain any subset of \(\{-1,0,1\}\) - including a crescent-shaped region enclosing both \(-1\) and \(1\) while excluding \(0\). By the residue theorem, \(\int_C f\,dz=2\pi i\sum(\text{residues enclosed})\). Listing all 8 subsets and their residue sums: \(\varnothing\to0\), \(\{0\}\to1\), \(\{1\}\to-\tfrac12\), \(\{-1\}\to-\tfrac12\), \(\{0,1\}\to\tfrac12\), \(\{0,-1\}\to\tfrac12\), \(\{1,-1\}\to-1\), \(\{0,1,-1\}\to0\).

Step 6: Multiplying each sum by \(2\pi i\) gives the values \(0,\ 2\pi i,\ -\pi i,\ -\pi i,\ \pi i,\ \pi i,\ -2\pi i,\ 0\), whose distinct set is \(\{0,\pm i\pi,\pm 2i\pi\}\).

\[\boxed{\{0,\ \pm i\pi,\ \pm 2i\pi\}}\]