Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2005 / 4

For a smooth curve $\gamma :[a,b]\to \mathbb{C}$ and continuous $f$ on its image, the contour integral is defined byThe derivative carries the direction and speed of traversal; the complex-valued real-parameter integral is defined by integrating its real and imaginary parts.

The chain rule gives $dF(\gamma (t))/dt=F^{\prime }(\gamma (t)){\gamma }^{\prime }(t)=f(\gamma (t)){\gamma }^{\prime }(t)$. The fundamental theorem of calculus therefore yieldsbecause the closed curve has equal endpoints. The existence of a single-valued holomorphic antiderivative on the whole neighborhood is the decisive hypothesis.

Take the unit circle $\gamma (t)=e^{it}$, $0\le t\le 2\pi$, in the punctured complex plane. Direct parameter integration givesA global holomorphic antiderivative would make this integral zero by the previous part. Thus there is no such antiderivative on the punctured plane. Local primitives can exist without combining into one single-valued function on this domain.

If a holomorphic function $g$ on the punctured complex plane satisfied $e^{g(z)}=z$, the chain rule would give $e^{g(z)}{g}^{\prime }(z)=1$, hence $g^{\prime }(z)=1/z$. This contradicts the previous part. Therefore there is no global holomorphic logarithm on the punctured plane. A branch of the complex logarithm requires a suitable smaller domain.