Question

Let \( C \) be a clockwise oriented closed curve in the complex plane defined by \( |z| = 1 \). Further, let \( f(z) = jz \) be a complex function, where \( j = \sqrt{-1} \). Then, \[ \oint_C f(z)\, dz = \underline{2cm} \quad \text{(round off to the nearest integer)}. \]

Hint:

If a function is analytic (holomorphic) inside and on a closed curve, the contour integral over that curve is zero. This is a direct result of {Cauchy's theorem}.

Answer 1

Given \( f(z) = jz \), this function is analytic (entire) everywhere in the complex plane. 

Since \( f(z) \) is analytic inside and on the closed contour \( C \), by Cauchy's theorem: \[ \oint_C f(z) \, dz = 0 \] Direction of traversal (clockwise or counter-clockwise) does not matter if the function is analytic over the region enclosed. \[ \boxed{0} \]