# 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?

• This is a poorly written problem. First, "slit" is not a conventional mathematical concept as used. Though the only sensible meaning I see is indeed $i[-2,2]$. Second, it says "domain", but what it describes is not a domain, which has to have interior points. Your answer is clearly analytic on $i[-2,2]$. The book's answer has the argument of $\text{Log}$ lying on the negative real axis when $z \in i[-2,2]$, which is its branch cut. It appears the author forgot their purpose and choose a function that makes $i[-2,2]$ the branch cut, not where it is analytic. Feb 23, 2020 at 20:26

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.