Question

The function \( f(Z)=\dfrac{1}{Z-1} \) of a complex variable \(Z\) is integrated on a closed contour in an anti-clockwise direction. For which of the following contours, does this integral have a non-zero value?

• A. \( |Z-2|=0.01 \)
• B. \( |Z-1|=0.1 \)
• C. \( |Z-3|=5 \)
• D. \( |Z|=2 \)

Hint:

- The contour integral \(\oint \frac{1}{z-a}\,dz\) equals \(2\pi i\) if and only if the contour encloses \(z=a\) (anticlockwise). Otherwise it is zero.

Answer 1

The Correct Option is B, C, D

Step 1: By Cauchy’s integral theorem, for \(f(Z)=\dfrac{1}{Z-1}\) the contour integral \(\displaystyle \oint f(Z)\,dZ\) is non–zero only if the contour encloses the simple pole at \(Z=1\); then \(\displaystyle \oint \frac{1}{Z-1}\,dZ = 2\pi i\) (for anticlockwise traversal).
Step 2: Check each contour:
- \( |Z-2|=0.01\): circle centered at \(2\) of radius \(0.01\) does \emph{not} enclose \(Z=1\) \(\Rightarrow\) integral \(=0\).
- \( |Z-1|=0.1\): circle centered at \(1\) of radius \(0.1\) encloses the pole \(\Rightarrow\) non–zero.
- \( |Z-3|=5\): circle centered at \(3\) of radius \(5\) also encloses \(Z=1\) \(\Rightarrow\) non–zero.
- \( |Z|=2\): circle centered at \(0\) of radius \(2\) encloses \(Z=1\) \(\Rightarrow\) non–zero.
Thus the integral is non-zero for options (B), (C) and (D).