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?

  • 1
    $\begingroup$ 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. $\endgroup$ Feb 23, 2020 at 20:26

1 Answer 1


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.

  • $\begingroup$ Oh, I'm so sorry. Hope you learn them soon! Also, maybe you should specified so in your question. :( $\endgroup$
    – EBO
    Feb 23, 2020 at 2:24

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