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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?

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I mistyped the equation in the title of the question...oops. –  Hanmyo Nov 27 '12 at 21:55

4 Answers 4

up vote 2 down vote accepted

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 $$ \sqrt{2z-\ln z -2}= h - \frac{1}{3} h^2 + \frac{7}{36} h^3 - \frac{73}{540} h^4 + \frac{1331}{12960} h^5 - \frac{22409}{272160} h^6 \pm\cdots) $$

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That is in fact the way I was supposed to do it...I was somehow confused about why $\sqrt{h^2}$ would be analytic, but I think I was thinking in terms of real values rather than $r\exp(i\theta)$ terms. Thank you for the epiphany. =) –  Hanmyo Nov 27 '12 at 22:33

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 $$ \sqrt{2z-2\log(z)-2}=(z-1)e^{\log(f(z))/2} $$

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Now here's some rigor I can get behind! –  Antonio Vargas Nov 28 '12 at 17:40

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

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I'm not sure how you see this. When my friend plotted the complex values of the function, $z=1$ didn't really look like a singularity. –  Hanmyo Nov 27 '12 at 22:20
2  
It is not, the logarithm does not have a singularity at $z=1$, it has one at $z=0$ (and at $z=\infty$). –  Lukas Geyer Nov 27 '12 at 23:28
    
Note that $$ \log(2z-2\log(z)-2)=2\log(z-1)-\frac23(z-1)+\frac5{18}(z-1)^2+\dots $$ so $\log(f(z))$ does have the same problem at $z=1$ that $\log(z-1)$ does. However, if we divide by $2$ and exponentiate, things work out. –  robjohn Nov 28 '12 at 18:22

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 $z=0$.

one-branch

  • 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...).

two-branches

Hoping this helped too,

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