Some Contour Integration Questions

[thatonekid]

Evaluate the contour integral. [FONT=MathJax_Main][FONT=MathJax_Size1]∫[FONT=MathJax_Math]c[/FONT][FONT=MathJax_Math] d[/FONT][FONT=MathJax_Math]z[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]c[/FONT][FONT=MathJax_Math]o[/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Math]z[/FONT][FONT=MathJax_Main])[/FONT][/FONT][/FONT]
(a) When c is the circle |z|=1 positively oriented;
(b) when c is the circle |z-pi| = 1.5 pi, positively oriented;
(c) when c is the circle |z-2i| = 1, positively oriented.

 

[Opalg]

Evaluate the contour integral. [FONT=MathJax_Main][FONT=MathJax_Size1]∫[FONT=MathJax_Math]c[/FONT][FONT=MathJax_Math] d[/FONT][FONT=MathJax_Math]z[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]c[/FONT][FONT=MathJax_Math]o[/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Math]z[/FONT][FONT=MathJax_Main])[/FONT][/FONT][/FONT]
(a) When c is the circle |z|=1 positively oriented;
(b) when c is the circle |z-pi| = 1.5 pi, positively oriented;
(c) when c is the circle |z-2i| = 1, positively oriented.

The starting point is that the integral of a meromorphic function round a positively oriented contour is equal to $2\pi i$ times the sum of the residues at all the poles inside the contour. In this case, the function is $\dfrac1{1-\cos z}$. So you need to find (i) where are the poles of this function, and (ii) what is the residue at each pole inside the contour? Of course, the poles inside the contour will differ for each of the three given contours.

For the answer to (i), the function will have a pole when the denominator is 0. So you need to think about when $\cos z = 1$. On the real axis, that happens when $z$ is a multiple of $2\pi$. Are those the only complex solutions of $\cos z = 1$, or are there other solutions off the real axis?

For (ii), what is the residue of $\dfrac1{1-\cos z}$ at a point where $\cos z = 1$? You will have to careful here, because the poles of this function are double poles. (That is because the power series for $\cos z$ starts $1-\frac12z^2+\ldots$, and so $1-\cos z$ has no term in $z$ but starts with a $z^2$ term.)

See how far you get with those hints, and come back here if you need further help.