Integrating over Branch Cuts I'm having problems following the solution for b). The main problem is finding the interval which you integrate over, which for some reason in this case is $(-i,i)$. To be frank I don't really get the rest of the answer either.


 A: The problem in the book is formulated in a lousy way. In the first place the curves $\gamma_{\rm a}$ and $\gamma_{\rm b}$ are not subsets of ${\mathbb C}$, but paths having a sense of direction, and secondly the notation $\bigl[{3\pi\over2},{\pi\over2}\bigr]$ is very bad in this context. At any rate, I had to peek at the suggested solution to understand what was meant.
So we have the two paths
$$\gamma_{\rm a}:\quad\theta\mapsto e^{i\theta}\quad\left(-{\pi\over2}\leq\theta\leq{\pi\over2}\right),\qquad \gamma_{\rm b}:\quad\theta\mapsto e^{-i\theta}\quad\left({\pi\over2}\leq\theta\leq{3\pi\over2}\right)\ .\tag{1}$$
With increasing $\theta$ both paths run  from $-i$ to $i$ along the unit circle, $\gamma_{\rm a}$ counterclockwise, and $\gamma_{\rm b}$ clockwise. Using the parametrizations $(1)$  we obtain
$$\int_{\gamma_{\rm a}}{dz\over z}=\int_{-\pi/2}^{\pi/2}{i\>e^{i\theta}\over e^{i\theta}}\>d\theta=i\pi,\qquad \int_{\gamma_{\rm b}}{dz\over z}=\int_{\pi/2}^{3\pi/2}{(-i)\>e^{-i\theta}\over e^{-i\theta}}\>d\theta=-i\pi\ .$$
The suggested solution proposes another way, namely using a primitive of the integrand (if we can find one). For $\gamma_{\rm a}$ this is easy, because the trace of $\gamma_{\rm a}$  is completely contained in the domain of ${\rm Log}$, i.e., of the principal value of the logarithm. Since ${\rm Log}$ is a primitive of the function $z\mapsto{1\over z}$ we can immediately write
$$\int_{\gamma_{\rm a}}{dz\over z}={\rm Log}(z)\Biggr|_{z=-i}^{z=i}=i\pi\ .$$
For $\gamma_{\rm b}$ more work is necessary. We first have to produce a primitive of $z\mapsto{1\over z}$ valid in a neighborhood of the trace of $\gamma_{\rm b}$. Such a primitive is constituted by the function
$$L(z):={\rm Log}(-z)\ ,$$
because the domain of $L$ is ${\mathbb C}\setminus\{{\rm positive\ real\ axis}\}$, and by the chain rule we have
$${d\over dz}L(z)={1\over -z}\cdot(-1)={1\over z}\ .$$
It follows that
$$\int_{\gamma_{\rm b}}{dz\over z}=L(z)\Biggr|_{z=-i}^{z=i}={\rm Log}(-i)-{\rm Log}\bigl(-(-i)\bigr)=-i\pi\ .$$
A: I agree with other people's remarks that the problem in the book is badly stated, but it appears to be this: evaluate
$$\int_\gamma\frac{dz}{z}\ ,$$
where the path $\gamma$ is defined by
$$z=e^{i\theta}\ ,\quad\hbox{$\theta$ from $3\pi/2$ to $\pi/2$},$$
by using an appropriate antiderivative (primitive function) of $1/z$.
As pointed out by yourself and others, in this case it is just as easy, if not more so, to parametrise the path.  However, if you had an integral like
$$\int_\gamma 3z^2\,dz$$
where $\gamma$ is given by
$$z=\frac{(3+4i)(t/4)^2\cos(\pi t/2)}{t^3+2it-500}\quad\hbox{for $t$ from $-5$ to $8$},$$
then parametrising the path would be a nightmare, but using an antiderivative can be done in your head!  So it is wise to know both methods, and presumably the point of the question is to give you practice in using antiderivatives.  The relevant result is:

if $f$ is continuous and has an antiderivative $F$ everywhere in a simply connected domain $D$, then
  $$\int_\gamma f(z)\,dz=F(\beta)-F(\alpha)$$
  for every path $\gamma$ which starts at $\alpha$, finishes at $\beta$, and lies entirely within $D$.

The first part of the book answer points out, in effect, that this theorem does not apply over the given contour for $f(z)=1/z$, with $F(z)$ being the principal value logarithm
$$F(z)=F(re^{i\theta})=(\ln r)+i\theta\quad
  \hbox{with $-\pi<\theta\le\pi$},$$
because $F$ is not analytic at the point $-1$ on $\gamma$, and therefore cannot be an antiderivative of $f$ at that point.
Instead, define a new logarithm function
$$F(z)=F(re^{i\theta})=(\ln r)+i\theta\quad
  \hbox{with $0\le\theta<2\pi$}.$$
It is possible to show that $F'(z)=1/z$ for all $z$ except those lying on the positive real axis (including the origin).  Therefore the above result can be used; the contour $\gamma$ goes from $-i$ (when $\theta=3\pi/2$) to $i$ (when $\theta=\pi/2$) and we have
$$\int_\gamma\frac{dz}{z}=F(i)-F(-i)
  =F(e^{i\pi/2})-F(e^{3i\pi/2})
  =\frac{i\pi}{2}-\frac{3i\pi}{2}=-i\pi\ .$$
A: The principle branch is defined as Log$z=\ln r+i\Theta$ with $r>0$ and $-\pi<\Theta <\pi$. So for the specified $\gamma$, since $\theta \in [3\pi/2,\pi/2]$ you can see that this interval "goes outside" the interval of the principle branch. So you must use another branch, and in this case $0 \leq \theta < 2\pi$ will "contain" $\theta \in [3\pi/2,\pi/2]$.
Now \begin{equation} z=e^{i3\pi/2}=\cos(3\pi/2)+i\sin(3\pi/2)=-i\end{equation} and \begin{equation} z=e^{i\pi/2}=\cos(\pi/2)+i\sin(\pi/2)=i,\end{equation} so there you have your integration interval. 
