Question

Let \( R = \{ z = x + iy \in \mathbb{C} : 0 < x < 1 \text{ and } - 11 \pi < y < 11 \pi \} \) and \( r \) be the positively oriented boundary of \( R \). Then the value of the integral \[ \frac{1}{2 \pi i} \int_r \frac{e^z}{e^z - 2} \, dz \] \text{is \(\underline{\hspace{1cm}}\) .}

Hint:

Use the residue theorem to evaluate contour integrals around simple poles, and find the residue by multiplying by the factor \( (z - \text{singularity}) \).

Answer 1

Correct Answer: 11

The given contour integral can be evaluated using the residue theorem. The integrand has a singularity where \( e^z = 2 \), which occurs when \( z = \ln 2 \). Thus, we need to calculate the residue of the function \( f(z) = \frac{e^z}{e^z - 2} \) at \( z = \ln 2 \). The residue of this function at \( z = \ln 2 \) is: \[ \text{Res}(f, \ln 2) = \lim_{z \to \ln 2} (z - \ln 2) \frac{e^z}{e^z - 2} = \lim_{z \to \ln 2} \frac{e^z}{1} = 2. \] By the residue theorem, the value of the contour integral is given by: \[ \frac{1}{2\pi i} \int_r \frac{e^z}{e^z - 2} \, dz = 2. \] Thus, the value of the integral is \( \boxed{2} \).