Question

Let \(C\) be a counter-clockwise circle defined as \(|z| = 2\). Then, the value of the complex integral \(\frac{1}{2\pi i} \oint_C \frac{z^2-1}{z^2+1}e^{-z} \, dz\) is _______

• A. \(\sin 1\)
• B. \(2 \sin 1\)
• C. \(\cos 1\)
• D. \(2 \cos 1\)

Hint:

Poles on the imaginary axis at \(\pm i\theta\) often simplify to trigonometric functions (\(\sin \theta\) or \(\cos \theta\)) when their residues are summed.
Always remember that \(\frac{1}{2\pi i}\) is already present in front of the integral, so the final answer is simply the sum of residues without multiplying by \(2\pi i\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the evaluation of a closed contour complex integral along a circle of radius 2 centered at the origin, traversed counter-clockwise.

Step 2: Key Formula or Approach:
According to Cauchy's Residue Theorem, if a function \(g(z)\) is analytic inside and on a simple closed contour \(C\) except at a finite number of isolated singular points \(z_k\) inside \(C\), then:
\[ \frac{1}{2\pi i} \oint_C g(z) \, dz = \sum \text{Residues of } g(z) \text{ at poles inside } C \] The residue of a function with a simple pole at \(z = z_0\) is calculated as:
\[ \text{Res}(g, z_0) = \lim_{z \to z_0} (z - z_0) g(z) \]

Step 3: Detailed Explanation:
Let us define the integrand function \(g(z)\):
\[ g(z) = \frac{z^2 - 1}{z^2 + 1} e^{-z} \] 1. Find the poles of \(g(z)\) by setting the denominator to zero:
\[ z^2 + 1 = 0 \implies z^2 = -1 \implies z = \pm i \] These are simple poles.
2. Determine which poles lie inside the contour \(C\) defined by \(|z| = 2\):
- For \(z_1 = i\): \(|z_1| = |i| = 1 \lt 2\), which lies inside \(C\).
- For \(z_2 = -i\): \(|z_2| = |-i| = 1 \lt 2\), which also lies inside \(C\).
Since both poles lie inside the circle of radius 2, we must compute the residues at both points.
3. Calculate the residue at \(z = i\):
\[ \text{Res}(g, i) = \lim_{z \to i} (z - i) \frac{z^2 - 1}{(z - i)(z + i)} e^{-z} \] \[ \text{Res}(g, i) = \frac{i^2 - 1}{i + i} e^{-i} = \frac{-1 - 1}{2i} e^{-i} = \frac{-2}{2i} e^{-i} = \frac{-1}{i} e^{-i} = i e^{-i} \] 4. Calculate the residue at \(z = -i\):
\[ \text{Res}(g, -i) = \lim_{z \to -i} (z + i) \frac{z^2 - 1}{(z - i)(z + i)} e^{-z} \] \[ \text{Res}(g, -i) = \frac{(-i)^2 - 1}{-i - i} e^{-(-i)} = \frac{-1 - 1}{-2i} e^{i} = \frac{-2}{-2i} e^{i} = \frac{1}{i} e^{i} = -i e^{i} \] 5. Sum the residues inside the contour:
\[ \sum \text{Residues} = \text{Res}(g, i) + \text{Res}(g, -i) = i e^{-i} - i e^{i} \] Factor out \(-i\) to match the standard definition of the sine function:
\[ \sum \text{Residues} = -i \left( e^{i} - e^{-i} \right) \] Recall the Euler formula definition for \(\sin(1)\):
\[ \sin 1 = \frac{e^{i} - e^{-i}}{2i} \implies e^{i} - e^{-i} = 2i \sin 1 \] Substitute this identity back into the residue sum:
\[ \sum \text{Residues} = -i \left( 2i \sin 1 \right) = -2 i^2 \sin 1 \] Since \(i^2 = -1\):
\[ \sum \text{Residues} = -2(-1) \sin 1 = 2 \sin 1 \] Applying Cauchy's Residue Theorem, the value of the integral is exactly this sum of residues.

Step 4: Final Answer
The value of the integral is \(2 \sin 1\), which corresponds to option (B).