Question

Compute the contour integral of f(z)=zⁿ for any n∈Z along three different paths γ+​,γ−​,,γ₀ going from z=+a to z=−a for a positive real number a. Namely the path γ+​halfcircling zero counterclockwise, the path γ- ​ half-circling zero clockwise and γ₀ going straight on the real axis. Careful, your calculation and result will depend at two places on n. (for γ₀ you may assume n≥0 so the integral is well-defined)

Answer 1

The contour integral of
`\( f(z) = z^n \)` along the path
`\( \gamma_+ \)` is (0), along the path
`\( \gamma_- \)` is (0), and along the path
`\( \gamma_0 \)` is
`\( (2\pi i)/(n+1) \) for \( n \\eq -1 \`). For
`\( n = -1 \)`, the integral along
`\( \gamma_0 \)` is undefined.

Step-by-step explanation:

When computing contour integrals of
`\( f(z) = z^n \)` along closed paths, we often utilize Cauchy's Residue Theorem. For
`\( \gamma_+ \)` and
`\( \gamma_- \)`, the paths involve half-circling zero counterclockwise and clockwise, respectively. The integral is (0) for both cases because the function
`\( f(z) = z^n \)` is analytic everywhere except at the origin, and there are no singularities enclosed by the paths.

For
`\( \gamma_0 \)`, which goes straight on the real axis from
`\( +a \)` to
`\( -a \)`, we apply the Residue Theorem to find the integral. The only singularity within the contour is at the origin, and the residue at
`\( z = 0 \)` is (0) for
`\( n \\eq -1 \)`. Therefore, the integral along
`\( \gamma_0 \)` is
`\( (2\pi i)/(n+1) \) for \( n \\eq -1 \)`. However, for
`\( n = -1 \)`, the integral is undefined, as it involves division by zero.

In summary, the contour integrals of
`\( f(z) = z^n \)` along the specified paths depend on the value of
`\( n \). For \( n \\eq -1 \)`, the integrals along
`\( \gamma_+ \`),
`\( \gamma_- \)`, and
`\( \gamma_0 \`) are all (0) except for
`\( \gamma_0 \), where it equals \( (2\pi i)/(n+1) \)`. For
`\( n = -1 \)`, the integral along \( \gamma_0 \) is undefined.