# Is this contour continuously deformable into a circle?

As an exam question, we had to solve the integral of $\frac{1}{z}$ over the following contour:

(The contour is a sequence of straights arcs joining -1, -$\frac{i}{2}$, $\frac{1}{2}$, i, $-\frac{1}{2}$, -i, 1, $\frac{i}{2}$ and -1).

I assumed for the time being that the contour was continuously deformable into a circle with radius 1 around the origin (by the Deformation Invariance Theorem) as the whole contour was around the origin as well. However, is it allowed to make this statement or is the only way to calculate it summing the integrals of the separate straight arcs?

• the curve is deformable to a circle which is twice covered. – James S. Cook Dec 17 '14 at 19:42

A curve which is contractible to the unit circle has necessarily winding number $1$, because curves with different winding numbers can not be continuously deformed into each other ("Are not homotopic", if you want to be more sophisticated).

Therefore, the answer is "No" because your curve has (as can be easily shown) a winding number of $2$.

Is this fine or do you need some more advice?

Edit:

We have only one pole inside our contour, so we can easily evaluate the integral by the Residue theorem: $$I=2 \pi i \times \text{winding number}\times \text{Res}[z=0]=2 \pi i \times2\times1=4\pi i$$

No need for the annoying calculations with the originally parametrized contour

• Thanks for the clarification. Would there have been another way to simplify this contour integral significantly or would the only realistic way be to parametrize each straight arc and calculate the integrals of each individual arc with help of the parametrization an? – surfer1311 Dec 17 '14 at 18:32
• See my edit above – tired Dec 17 '14 at 18:47
• hi, have you understand it now? – tired Dec 18 '14 at 12:31