• Say I have the function $\log (f(z))$, then does the imaginary part of the value go down by $-i 2 \pi$ on crossing the branch-cut of the log function even if $f(z)$ is not crossing the branch-cut?

  • Specifically, consider the function $\log(z^2 +a ^2)$ where $a \in \mathbb{R}$. And I am choosing the branchcuts to be going up/down along the positive/negative imaginary axis starting at $\pm ia$. So if you hit the positive imaginary axis above $ia$ from the right then the imaginary part of the logarithm would be $i\pi$ whereas from the left would be $-i\pi$. I guess that its the same below $-ia$ i.e approach from the right leads to $i\pi$ whereas from the left it leads to $-i\pi$.

Now I am taking a circle around the point $ib$ (say $b>a>0$ or you can also help with the similar case in the LHP)parameterized as $z = ib + \epsilon e^{i\phi}$. Now here begins the confusion -

(1) naively I would have thought that I have to add "by hand" a $-i2\pi$. To the $\log$ when $ -\pi < \phi < -\pi/2$ and $ \pi < \phi < \pi/2$ (i.e when the $z$ is to the left of the branch-cut)

(2) But then in the needed limit of $\epsilon \rightarrow 0$ as $z$ goes around this circle the actual function that the $\log$ sees i.e $z^2 + a^2$ is going around an infinitesimal circle with its center at $b^2-a^2$. So the function $z^2 + a^2$ is far away from the branch-cuts and the branch-points and is not crossing them. So by this argument I would think that no phase adjustments need to be done for the $\log$ as $z$ sweeps out the circle around $ib$.

So which of these (1) or (2) is the right way to think?

  • $\begingroup$ If $z = ib$, then $z^{2} + a^{2} = -b^{2} + a^{2} < 0$, not $a^{2} - b^{2} > 0$, so even a small circle crosses the branch cut. That is, your reasoning in (2) is incorrect. (IIRC, below $-ia$, the imaginary part of $\log(z^{2} + a^{2})$ approaches $-\pi i$ from the right and $i\pi$ from the left. To see this, think about the effect of squaring, and what happens when polar angles near the cut are doubled.) $\endgroup$ Commented Apr 17, 2014 at 22:46
  • $\begingroup$ @user86418 I am not getting you - in reasoning (2) the radius of the circle is $O(\epsilon)$ and its center is somewhere on the $x$ axis - then how is it going to hit any of the branch-cuts? (the branchcuts are a finite distance $a$ away from the x-axis) - so are you saying that the phase of the "log" is determined by the location of $z$ and not by the location of its argument i.e $z^2 + a^2$? $\endgroup$
    – user6818
    Commented Apr 17, 2014 at 23:19
  • $\begingroup$ @user86418 It would be helpful if you can elaborate on how these phases $i\pi$ and $-i\pi$ determined. I am kind of confused. [what you are saying surely right - I checked it on Mathematica] $\endgroup$
    – user6818
    Commented Apr 17, 2014 at 23:35

1 Answer 1


Let's agree explicitly that "$\log$" refers to the branch of logarithm defined on $\mathbf{C}\setminus(-\infty, 0]$ whose imaginary part is between $-\pi i$ and $\pi i$, and that "$f(z) \leq a$" means "both $a$ and $f(z)$ are real, and $f(z) \leq a$".

If $f$ is holomorphic in some region $U$, then $\log(f)$ is holomorphic at $z$ in $U$ provided $f(z)$ does not lie on the branch cut, i.e., "$f(z) \leq 0$" is false.

In your example, $f(z) = z^{2} + a^{2}$ is holomorphic on all of $\mathbf{C}$, so $\log f$ is holomorphic on the twice-cut plane, from which the imaginary intervals $i(-\infty, -a]$ and $i[a, \infty)$ have been removed. These intervals are precisely the set of $z$ for which $z^{2} + a^{2} \leq 0$. In fancier language, they constitute the preimage of the cut $(-\infty, 0]$ under $f$.

The issue in (2) is that a small circle $C$ centered at $ib$ (with $|b| > a$) crosses the "cuts" of $\log(f)$. The image of $C$ under $f$ is a closed curve (not a circle) "centered on" the negative real axis, so $f(C)$ crosses the branch cut of $\log$. Locating $-b^{2} + a^{2}$ in the $z$ plane, relative to the branch cuts of $\log(f)$, is immaterial: We're not evaluating $\log(f)$ near $-b^{2} + a^{2}$, we're evaluating $\log$ itself. $\ddot\smile$

To see what's going on with the values of $\log(f)$ "near" each imaginary cut, consider points $z = re^{i\phi}$ with $r > a$ and $\phi$ "close to" $\pm\pi/2$: $$ z^{2} + a^{2} = r^{2} e^{2i\phi} + a^{2} = (r^{2} \cos(2\phi) + a^{2}) + ir^{2} \sin(2\phi). $$ If $\phi$ is a bit smaller than $\pm\pi/2$, then $\sin(2\phi) > 0$. Consequently, $z^{2} + a^{2}$ has "small" positive imaginary part, so $\log(z^{2} + a^{2})$ has imaginary part close to $\pi i$.

Similarly, if $\phi$ is a bit larger than $\pm\pi/2$, then $\sin(2\phi) < 0$ and $\log(z^{2} + a^{2})$ has imaginary part close to $-\pi i$.

  • $\begingroup$ Thanks for the explanations! I guess for the $\pm i\pi$ argument of yours you need to also note that for all the cases $a^2 + r^2 Cos(2\pi)$ is negative - right? $\endgroup$
    – user6818
    Commented Apr 20, 2014 at 4:16
  • $\begingroup$ So to summarize I guess one can say that when one is going around in a circle about a point $i(b>a>0)$ or $-ib$ as $z=ib + \epsilon e^{i\phi}$ then for $-\frac{\pi}{2} \leq \phi \leq \frac{\pi}{2}$ lets call it $log_R(z^2+a^2)$ and for other values of $\phi$ lets call it $log_L(z^2+a^2)$. Then (1) in the UHP we have $log_L (z^2 + a^2) = log_R (z^2+a^2) -2 i \pi$ and (2) in the LHP we have $log_L (z^2 + a^2) = log_R (z^2+a^2) +2 i \pi$ - right? Thats how one would differentiate between the two things in doing a contour integral - right? $\endgroup$
    – user6818
    Commented Apr 20, 2014 at 4:23
  • $\begingroup$ Also I guess that when one is going about such a similar circle around $\pm ia$ then the difference between $log_L$ and $log_R$ needs to be made only for $0 \leq \phi \leq \pi$ - right? And for the other-half of the circle the $log$ is the same on both the sides - right? $\endgroup$
    – user6818
    Commented Apr 20, 2014 at 4:27
  • $\begingroup$ And the absolute assignments of $\pm ia$ never matter - right? $\endgroup$
    – user6818
    Commented Apr 20, 2014 at 4:28
  • $\begingroup$ As you say, $a^{2} + r^{2}\cos(2\phi) < 0$ when $\phi$ is "close to" $\pm\pi/2$. (This is implicit in my "preimage of the cut" remark.) I'm less certain about your remaining comments: There's only one "$\log$" function in this entire discussion; the function $z \mapsto \log(z^{2} + a^{2})$ is defined everywhere except the two imaginary cuts. If you try to use $\log_{L}$ and $\log_{R}$ to indicate limiting behaviors near the cuts, you can't compare $\log_{L}(z^{2} + a^{2})$ and $\log_{R}(z^{2} + a^{2})$, since their domains don't overlap. $\endgroup$ Commented Apr 20, 2014 at 12:22

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