Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2010 / 5

The complex line integral along the smooth parametrized curve isFor the reversed curve, the chain rule gives ${\tilde{\gamma }}^{\prime }(t)=-{\gamma }^{\prime }(a+b-t)$. Substitution $s=a+b-t$ yieldsConsequently reversing orientation negates the complex line integral.

An entire antiderivative is $F(z)=z\sin z+\cos z$, since differentiation cancels the sine terms and leaves $F^{\prime }(z)=z\cos z$. The curve runs from $-i$ to $i$. Both terms of $F$ are even, so $F(i)=F(-i)$, and the complex line integral gives

Along the circle, $z=z_0+r{e}^{it}$ and $dz=ir{e}^{it}dt$. Expanding the nonholomorphic integrand givesAfter multiplication by $ir{e}^{it}$, all nonconstant integer-frequency exponentials have zero integral over $[0,2\pi ]$. The remaining constant term is $i{r}^2{z}_0$. ThereforeThis direct parametrization avoids applying a holomorphic-antiderivative argument to $|z{|}^2$, which is not holomorphic.