The value of the integral ∫eᶻ-1/e dz along the unit circle is 2πi.
We can evaluate the integral ∫eᶻ-1/e dz along the unit circle using the Cauchy integral formula.
Here's the approach:
Identify the singularities:
The integrand eᶻ-1/e has a singularity at z = 0. Since the unit circle encloses z = 0, we need to evaluate the residue of the integrand at this point.
Residue theorem:
The Cauchy integral formula states that for a function f(z), analytic within and on a simple closed contour C except for a finite number of poles a1, a2, ..., an inside C, where f(z) is also finite, we have:
∫C f(z) dz = 2πi ∑ni=1 Res(f, ai)
where Res(f, ai) is the residue of f(z) at the pole a_i.
Evaluate the residue:
The residue of eᶻ-1/e at z = 0 is 1. This can be found using various methods, such as Laurent series expansion or direct differentiation.
Apply the Cauchy integral formula:
Using the residue theorem, we get:
∫eᶻ-1/e dz = 2πi * Res(eᶻ-1/e, 0) = 2πi * 1 = 2πi
Therefore, the value of the integral ∫eᶻ-1/e dz along the unit circle is 2πi.