Help Finding a branch of the function analytic in the given domain Having trouble finding the branch sets for given analytic functions. Is my procedure correct or am I confused?
Find a branch of the function analytic in the given domain
$(4+z^2)^{1/2}$ in the complex slit along the imaginary axis from $-2i$ to $2i$ 
Attempt: find where the function is negative real and zero.
This occurs when $4+z^2=-x,x\geq 0$,
This occurs when $z=i\sqrt{x+4}$ or $x=-i\sqrt{x+4}$
Thus the branch cut appears to occur along the imaginary axis from $2i$ to infinity, and from negative infinity to $-2i$ It appears we can use the principal branch for the function $e^{1/2Log(4+z^2)}$ here.
But the book says we must use $ze^{1/2Log(\frac{4}{z^2}+1)}$
Why cannot we use the principal branch mentioned?
 A: Just to make sure we're in the same page... By a branch of the logarithm in a region $D$ we understand an analytic function $g$ in $D$ which satisfies $\exp(g(z))=z$ for every $z\in D$. When we're asked to find a branch cut, we're asked to find the region $D$ and the function $g$.

Now, consiredering your problem...
You can see it this way. If we construct a branch for the logarithm for $z^2+4$ then we're done: the branch for the square root will be 
$$
  \exp (1/2 \times\text{the branch of the logarithm we constructed})
  .
$$
Suppose we have such branch of the logarithm in our region of interest $D$. Call it $\log$. Our branch of the logarithm must satisfy
$$
  \log (z^2+4)' = \frac{2z}{z^2+4}\quad\forall\:z\in D
  ,
$$
in other words
$$
  \log(z^2+4)=\int_{\gamma_z}\frac{2\zeta\,d\zeta}{\zeta^2+4}\quad\forall\:z\in D
  ,
$$
where $\gamma_z$ is any curve connecting some privileged (distinguished) point $z_0\in D$ with $z\in D$ which is contained in $D$. Then the problem arises: we need to ensure that such integral is well defined. This is where our region $D$ comes to consideration. What we try to do next is justify the well-definess of the integral above (it will depend on the region $D$ as we pointed out). Once it is well defined, we define the branch of the desired logarithm as the integral we proved is well defined and we're done.
If $D=\mathbb{C}\setminus\{z\in\mathbb{C}\colon\ \mathrm{Im}(z)\geq 2\ \text{or}
    \ \mathrm{Im}(z)\leq 2\}$. (The region you're sugesting.) Then we're done because
no matter where our curve $\gamma_z$ lies in $D$ we'll never have to worry about the bad points $2i$ and $-2i$. To prove that the integral does not depend on the given curve $\gamma_z$ we consider another one, call it $\sigma_z$, and consider the closed curve $\gamma_z-\sigma_z$. Such curve will not contain the bad points. Inside this curve the given function $2\zeta/(\zeta^2+4)$ is analytic and therefore the integral over this curve is zero and we obtain the equality of the integrals over $\gamma_z$ and $\sigma_z$.
Then, we've constructed a branch of the logarithm for this particular $D$. This answer is correct.
Now, if we consider $G=\mathbb{C}\setminus\{z\in\mathbb{C}\colon\ -2\leq\mathrm{Im}\leq 2\}$, can you prove that this integral is well defined? Detailed hint: Consider two different curves $\gamma_z$ and $\sigma_z$ and their difference $\Gamma_z$. Remember that both $\gamma_z$ and $\sigma_z$ are contained in $G$, so that the closed curve $\Gamma$ is also contained in $G$. Use the residue theorem if necessary.
