Question

The value of the integral $ \int_C \frac{e^{2z}}{(z-1)(z-2)}dz $ where C is the circle $ |z|=3 $ is

• A. \( \pi i(e^2-e^4) \)
• B. \( 2\pi i(e^4-e^2) \)
• C. \( \pi i(e^4-e^2) \)
• D. \( 2\pi i(e^2-e^4) \)

Hint:

For complex integrals, first identify all singularities. Then, check which singularities lie inside the given contour. If there are singularities inside, use the Residue Theorem: \( \oint_C f(z) dz = 2\pi i \sum \text{Res}(f, z_k) \). For a simple pole \( z_0 \), \( \text{Res}(f, z_0) = \lim_{z \to z_0} (z-z_0) f(z) \).

Answer 1

The Correct Option is B

Step 1: Identify the function and the contour.
The given integral is \( \oint_C \frac{e^{2z}}{(z-1)(z-2)}dz \).
The contour C is the circle \( |z|=3 \). This circle is centered at the origin with a radius of 3.
Step 2: Identify the singularities of the integrand.
The integrand is \( f(z) = \frac{e^{2z}}{(z-1)(z-2)} \).
The singularities (poles) of the integrand are where the denominator is zero, i.e., \( (z-1)(z-2) = 0 \).
This gives us \( z=1 \) and \( z=2 \).
Step 3: Determine which singularities lie inside the contour C.
For \( z=1 \), \( |1| = 1 \). Since \( 1<3 \), \( z=1 \) lies inside the circle \( |z|=3 \).
For \( z=2 \), \( |2| = 2 \). Since \( 2<3 \), \( z=2 \) also lies inside the circle \( |z|=3 \).
Since both singularities lie inside the contour, we will use the Residue Theorem. The Residue Theorem states that if \( f(z) \) has a finite number of isolated singularities inside a simple closed contour C, then \( \oint_C f(z) dz = 2\pi i \sum \text{Res}(f, z_k) \), where the sum is over all singularities \( z_k \) inside C.
Step 4: Calculate the residues at each pole.
Both \( z=1 \) and \( z=2 \) are simple poles.
The residue at a simple pole \( z_0 \) for a function of the form \( f(z) = \frac{\phi(z)}{z-z_0} \) where \( \phi(z_0) \neq 0 \) is given by \( \text{Res}(f, z_0) = \phi(z_0) \).
For \( z=1 \): Let \( \phi(z) = \frac{e^{2z}}{z-2} \). \[ \text{Res}(f, 1) = \phi(1) = \frac{e^{2(1)}}{1-2} = \frac{e^2}{-1} = -e^2 \] For \( z=2 \): Let \( \phi(z) = \frac{e^{2z}}{z-1} \). \[ \text{Res}(f, 2) = \phi(2) = \frac{e^{2(2)}}{2-1} = \frac{e^4}{1} = e^4 \]
Step 5: Apply the Residue Theorem.
The sum of the residues is \( \sum \text{Res}(f, z_k) = \text{Res}(f, 1) + \text{Res}(f, 2) = -e^2 + e^4 = e^4 - e^2 \).
According to the Residue Theorem: \[ \oint_C f(z) dz = 2\pi i \sum \text{Res}(f, z_k) \] \[ \oint_C \frac{e^{2z}}{(z-1)(z-2)}dz = 2\pi i (e^4 - e^2) \] The final answer is $\boxed{\text{2}}$.