Why is oint_|z|=1(dz)/(z)≠ 0 even though 1/z is holomorphic on ℂsetminus\0\?

Because 1/z has no primitive on ℂsetminus\0\

Because the circle is not a contour

Because 1/z has a primitive but it is multivalued at the origin

Exercise · ZW-COMPLEX-PRIMITIVE-007 · ★★☆ Intermediate

Why is \(\oint_{|z|=1}\frac{dz}{z}\neq 0\) even though \(1/z\) is holomorphic on \(\mathbb{C}\setminus\{0\}\)?

Choose one option.

Show answer

D. Because \(1/z\) has no primitive on \(\mathbb{C}\setminus\{0\}\)

Explanation

If there were a primitive \(F\) on all of \(\mathbb{C}\setminus\{0\}\), the closed integral would be \(F(b)-F(a)=0\). The logarithm works only on regions that do not surround the origin.

Study Complex analysis in ZWMath →

Review the concept: Complex primitive

A primitive of a function \(f\) on an open set \(D\) is a holomorphic function \(F\) on \(D\) whose derivative is \(f\) at every point.

\[\int_{z_0}^{z_1} f(z)\,dz=F(z_1)-F(z_0)\]

Why is \(\oint_{|z|=1}\frac{dz}{z}\neq 0\) even though \(1/z\) is holomorphic on \(\mathbb{C}\setminus\{0\}\)?

Explanation

Review the concept: Complex primitive

More practice problems: complex primitive