Proving $\sqrt{2z-2\log(z)-2}$ is analytic near $z=1$.

Question

I am trying to prove $f(z)=\sqrt{2z-2\log(z)-2}$ is analytic near $z=1$. The issue is proving there is no branch point.

If I try the approach of taking the path $z=1+r\exp(i\theta)$ with $r=\epsilon$ and $\theta$ varying from $0$ to $2\pi$, I'm finding it hard to show that the value did not change:

For $f(r,\theta)$, $f(\epsilon,0) = \sqrt{2r\exp(i0)-2\log(1+r\exp(i0))} = \sqrt{2r-2\log(1+r)}$ and $f(\epsilon,2\pi) = \sqrt{2r\exp(i2\pi)-2\log(1+r\exp(i2\pi))} = \,\,??$.

Not sure if this is the right approach, but it's how I learned to do it. Any advice?

Answer 1

With $z=1+h$, we have $$\begin{array}2z-2\ln z -2&=2h-2\ln(1+h)\\& = 2h-2h+h^2-\frac23h^3+\frac12h^4-\frac25h^5\pm\cdots\\ &=h^2\cdot(1-\frac23 h+\frac12 h^2-\frac25h^3\pm\cdots)\end{array}$$ The square root of the second factor is analytic near $h=0$, and of course the first factor yields simply $h$.

(By the way, we find the expansion starting $$ \sqrt{2z-\ln z -2}= h - \frac{1}{3} h^2 + \frac{7}{36} h^3 - \frac{73}{540} h^4 + \frac{1331}{12960} h^5 - \frac{22409}{272160} h^6 \pm\cdots) $$

Answer 2

$z=1$ is a logarithmic branch point of the function $ f(z)=\sqrt{2z-2\log(z)-2}.$

Answer 3

As a complement let's show two pictures of the imaginary part of $f(z)$ :

• the first one shows the plot of $\sqrt{2z-2\log(z)-2}\ $ ($z=1$ is at the crossing point of the two curve (at the top). We may note the logarithmic branch point at $z=0$.

• the second one shows the plot of $\sqrt{2z-2\log(z)-2}\ $ and $-\sqrt{2z-2\log(z)-2}$ superposed ($z=1$ is at the crossing point of the two curve (at the front since the point of view is opposed)

  Note that we could simply have chosen for the two branches the two sheets crossing at $z=1$ with nothing special else (instead of the separation by colors from the software...).

Hoping this helped too,

Answer 4

Inside the circle $|z-1|=1$, $\log(z)$ is well-defined by $$ \log(z)=\int_1^z\frac{\mathrm{d}w}{w} $$ where the integral is taken over any path from $1$ to $z$ that stays inside the circle. Any two such paths give the same value since their difference is a loop that does not encompass the singularity at $z=0$.

Note that $$ \displaystyle\lim_{z\to1}\frac{2z-2\log(z)-2}{(z-1)^2} =\lim_{z\to1}\frac{2-2/z}{2(z-1)} =\lim_{z\to1}\frac{2/z^2}{2} =1 $$ Thus, $\displaystyle f(z)=\frac{2z-2\log(z)-2}{(z-1)^2}$ has a removable singularity at $z=1$, and does not vanish in some neighborhood, $\mathcal{N}$, of $z=1$. For $z\in\mathcal{N}$, we can define $$ \log(f(z))=\int_1^z\frac{f'(w)}{f(w)}\,\mathrm{d}w $$ where the integral is taken over any path contained in $\mathcal{N}$. Therefore, for $z\in\mathcal{N}$, we can define $$ \sqrt{2z-2\log(z)-2}=(z-1)e^{\log(f(z))/2} $$