Find a branch for $(4+z^2)^{1/2}$ such that it is analytic in the complex plane slit along the imaginary axis from $-2i$ to $2i$ Find a branch for the multiple-valued function $(4+z^2)^{1/2}$ such that it is analytic in the complex plane slit along the imaginary axis from $-2i$ to $2i$
Also, isn't this function already analytic on the slit from $-2i$ to $2i$ without being modified?
 A: The trick with these radicals of polynomials is usually to factor them into linear factors and choose a branch of the logarithm for every factor such that cancellation occurs on some of the branch cuts. In this case the process is particularly simple and it suffices to choose one branch of the logarithm for both factors, which is the one that has the branch cut along the negative imaginary axis
$$ -\frac{1}{2} \pi \le \arg \operatorname{Log} z < \frac{3}{2}\pi.$$
We will continue to use the lower case log for the branch with the cut along the negative real axis.
Now we define 
$$ f(z) = \exp(1/2 \operatorname{Log}(z-2i)) \exp(1/2 \operatorname{Log}(z+2i)).$$
The problem requires us to do two things: show that $f(z)$ is analytic in the complex plane slit between the two logarithmic singularities through the origin and verify that it agrees with $\sqrt{x^2+4}$ on the real line.
For the first part we certainly know that $f(z)$ is analytic in the complex plane slit from $2i$ to infinity along the negative imaginary axis. But we can show that it is in fact continuous across the slit below $-2i.$ To do this consider the point $z=-ti$ with $t>2$ and its two neighbors: $-ti-\epsilon$ in the left half plane and $-ti+\epsilon$ in the right half plane, with $\epsilon$ real and going to zero. For the left point we have
$$ f(z) = \exp(1/2 \log(t+2) + 1/2 \times 3/2 \pi i)
\exp(1/2 \log(t-2) + 1/2 \times 3/2 \pi i) \\=
\sqrt{t^2-4} \exp(+3/2 \pi i).$$
For the right point we get
$$ f(z) = \exp(1/2 \log(t+2) - 1/2 \times 1/2 \pi i)
\exp(1/2 \log(t-2) - 1/2 \times 1/2 \pi i) \\=
\sqrt{t^2-4} \exp(- 1/2 \pi i).$$
But since $\exp(- 1/2 \pi i) = \exp(+ 3/2 \pi i) = -i$ these two values are the same, so we have continuity across the cut. To get analyticity, apply Morera's theorem to a circle lying on the cut. Split it along a left half and a right half each consisting of an arc whose endpoints are connected by a line and integrate along both halves in counterclockwise direction. The two line segments cancel by continuity across the cut. The integral along the left segment is zero, as is the one along the right segment. Hence the integral along the circle lying on the cut is zero, and by Morera's theorem $f(z)$ is analytic there.
We still need to show that $f(z)$ agrees with $\sqrt{x^2+4}$ on the real axis. We will do the case $x>0.$ By Pythagoras the modulus of $x-2i$ is $\sqrt{x^2+4}$ as is the modulus of $x+2i.$ The point $x-2i$ forms some angle with the $x$ axis, call it $\theta$. The salient feature here is that $-1/2\pi < \theta < 0$ and that the point $x+2i$ forms that same angle with the $x$ axis, only positive, and there is no discontinuity in the angle because these two are entirely contained in the range of the argument of $\operatorname{Log} z.$ It follows that
$$f(z) = \exp(1/2 \log\sqrt{x^2+4} + 1/2i\theta) \exp(1/2 \log\sqrt{x^2+4} - 1/2i\theta) \\=\exp(1/2 \log\sqrt{x^2+4}) \exp(1/2 \log\sqrt{x^2+4}) = \exp(\log\sqrt{x^2+4}) =
\sqrt{x^2+4}.$$
A: The usual square root function needs a cut that comes in from $\infty$ to $0$ because there are two complex numbers that are the square root of any given number.  This is often taken as the negative real axis.  Thenfor real $a$ and small real $\epsilon,  (a+i\epsilon)^{\frac 12}$ is about $i\sqrt a$ and $(a-i\epsilon)^{\frac 12}$ is about $-i\sqrt a.  $You need to find an expression that for any $z$'s that are near (as long as they are not on opposite sides of this segment) the function values are near as well.
