M3M6: Methods of Mathematical Physics

$$ \def\dashint{{\int\!\!\!\!\!\!-\,}} \def\infdashint{\dashint_{\!\!\!-\infty}^{\,\infty}} \def\D{\,{\rm d}} \def\dx{\D x} \def\C{{\mathbb C}} \def\CC{{\cal C}} \def\HH{{\cal H}} \def\I{{\rm i}} \def\qqfor{\qquad\hbox{for}\qquad} $$

Dr. Sheehan Olver
s.olver@imperial.ac.uk

Office Hours: 3-4pm Mondays, Huxley 6M40
Website: https://github.com/dlfivefifty/M3M6LectureNotes

Lecture 8: Functions with branch cuts

We now discuss functions with branch cuts

1. $\log z$ with a cut on $(-\infty,0]$
2. $z^\alpha$ with a cut on $(-\infty,0]$
3. $\sqrt{z-1}\sqrt{z+1}$ with a cut on $[-1,1]$
4. $\rm{sign}(|z|-1)$ with a cut on the unit circle.

This is a step towards a Cauchy integral theorem on cuts, for recovering a holomorphic function from its behaviour on a cut.

1. Complex logarithm
2. Algebraic powers
3. Inferred analyticity

We begin with $\log z$.

Complex logarithm

Definition (Complex Logarithm) $$\log z = \int_1^z {1 \over \zeta} d\zeta$$ where the integral is understood to be on a straight line segment, that is $$\log z = \int_{\gamma_z} {1 \over \zeta} d\zeta$$ where $\gamma_z(t) = 1 + (z-1)t$ for $0 \leq t \leq 1$.

Demonstration this shows the integral path for a point $z$. We see how the path avoids the pole of $\zeta^{-1}$ at the origin:

This is well-defined apart from $z \in (-\infty,0]$, where there is a pole on the contour. This induces a branch cut: a jump in the value of the function.

We see that the limits from above and below exist: we can define $$ \begin{align*} \log_+ x &= \lim_{\epsilon \rightarrow 0^+} \log(x+\I \epsilon) \\ \log_- x &= \lim_{\epsilon \rightarrow 0^+} \log(x-\I \epsilon) \end{align*} $$ and calculate their values via:

By deformation, the value of the integrals is independent of the path: here we calculate the integral on an arc:

This works all the way to the negative real axis. Here we calculate $\log_\pm 1$ using integrals over half circles:

Combining the two contours we have $$ \log_+(-1) - \log_-(-1) = \oint {\D\zeta \over \zeta} = 2 \pi \I $$

We can establish some properties. First we show that $\log z = - \log {1 \over z}$ by considering the change of variables $\zeta = {1 \over s}$. Because $\gamma_z(t)^{-1}$ stays uniformly in the lower-half plane, we can deform it to a straight contour, which gives us the result: $$ \log z = \int_{\gamma_z} {\D\zeta \over \zeta} = -\int_{{1 \over \zeta} \circ \gamma_z} {\D s \over s} = - \int_{\gamma_{z^{-1}}} {\D s \over s} = -\log z^{-1} $$

Here's a plot of the relevant contours:

Here we see by looking at the phase plot that the two functions match:

Algebraic powers

Definition (algebraic power) $$z^\alpha = e^{\alpha \log z}$$

Note, for example, when $\alpha = 1/2$, $\sqrt z \equiv z^{1/2}$ is only one solution to $y^2 = z$.

Here's a phase plot showing that $\sqrt{z}$ also has a branch cut on $(-\infty,0]$:

On the branch cut along $(-\infty,0]$ it has the jump:

$${x_+^\alpha \over x_-^\alpha} = e^{\alpha (\log_+ x - \log_- x)} = e^{2 \pi \I \alpha}$$

In particular,

$$\sqrt{x}_+ = -\sqrt{x}_- = i \sqrt{|x|}$$

These are multiplicative jumps. We also have a subtractive jumps:

\begin{align*} x_+^\alpha - x_-^\alpha &= e^{\alpha \log_+ x} - e^{\alpha \log_- x} = e^{\alpha \log(-x) + \I \pi \alpha} - e^{\alpha \log(-x) - \I \pi \alpha} \cr & = 2 \I (-x)^\alpha \sin \pi \alpha \end{align*}

and an additive jump: \begin{align*} x_+^\alpha + x_-^\alpha & = 2 (-x)^\alpha \cos \pi \alpha \end{align*}

In particular, for $x < 0$, \begin{align*} \sqrt{x}_+ - \sqrt{x}_- &= 2 \I \sqrt{-x} \\ \sqrt{x}_+ + \sqrt{x}_- &= 0 \end{align*}

Let's look at another example: $\varphi(z) = \sqrt{z-1}\sqrt{z+1}$.

For $x > 0$ it's holomorphic. For $-1 < x < 1$ we have the multiplicative jump

$$\varphi_+(x) = \sqrt{x-1}_+ \sqrt{x+1} = - \sqrt{x-1}_- \sqrt{x+1} = -\varphi_-(x)$$

which gives the additive jump $$ \varphi_+(x) + \varphi_-(x) = 0 $$

But we also have a subtractive jump: $$ \varphi_+(x) - \varphi_-(x) = (\sqrt{x-1}_+ - \sqrt{x-1}_-) \sqrt{x+1} = 2\I \sqrt{1-x}\sqrt{x+1} = 2\I \sqrt{1 - x^2} $$

For $x < -1$ we actually have continuity:

$$\varphi_+(x) = \sqrt{x-1}_+ \sqrt{x+1}_+ = (- \sqrt{x-1}_-)(- \sqrt{x+1}_-) = \varphi_-(x)$$

Inferred analyticity

But continuity implies analyticity, using Cauchy's integral formula:

Theorem (continuity on a curve implies analyticity) Let $D$ be a domain and $\gamma \subset D$ a contour. Suppose $f$ is analytic in $D \backslash \gamma$, and continuous on the interior of $\gamma$. Then $f$ is analytic in $D \backslash \{\gamma(a), \gamma(b) \}$.

Sketch of Proof

For simplicity, suppose $D$ is a circle of radius 2 and $\gamma$ is the interval $[-1,1]$. For any $z$ off the interval, we can write $$ f(z) = {1 \over 2 \pi \I} \oint_{\Gamma_x} {f(\zeta) \over \zeta - z} \D\zeta $$ where $\Gamma_x$ is a simple closed contour that surrounds $z$ and passes through $x$ in both directions:

Because $f$ is continuous at $x$, we have $$f_+(x) = f_-(x) = f(x)$$ where $$f_\pm(x) = \lim_{\epsilon \rightarrow 0} f(x \pm \I \epsilon) $$ Therefore, the two integrals along $[-1,1]$ cancel out and we get: $$ f(z) = {1 \over 2 \pi \I} \oint_{{\tilde \Gamma}_x} {f(\zeta) \over \zeta - z} \D \zeta $$ where ${\tilde \Gamma}_x$ is $\Gamma_x$ with the contour on the interval removed:

This integral expression holds for all $z$ inside the contour $\tilde \Gamma_x$ but off the interval. But it therefore holds true for $f(x) = f_+(x) = f_-(x)$ by taking limits. Thus $$f(x) = {1 \over 2 \pi \I} \int_{{\tilde \Gamma}_x} {f(\zeta) \over \zeta -x} \D\zeta$$ hence $f$ is analytic at $x$.

⬛️

Theorem (weaker than pole singularity implies analyticity) Suppose $f$ is analytic in $D \backslash \{ z_0 \}$ and has a weaker than pole singularity at $z_0$: $$\lim_{z \rightarrow z_0} (z-z_0) f(z) = 0$$ holds uniformly. Then $f$ is analytic at $z_0$. (More precisely: $f$ can be analytically continued to $z_0$.)

Proof

Around $z_0$ is an annulus $A_{R0}$ inside which $f$ is analytic. Consider $z$ in $A_{R0}$ and a positively oriented circle $\gamma_r$ of radius $r$ with $|r| < |z-z_0|$. Then we have $$ f(z) = {1 \over 2 \pi \I} \oint_{\gamma_R} {f(\zeta) \over \zeta - z} \D \zeta - {1 \over 2 \pi \I} \oint_{\gamma_r} {f(\zeta) \over \zeta - z} \D \zeta . $$ here's a plot:

But we have $$ \left| \oint_{\gamma_r} {f(\zeta) \over \zeta - z} \D \zeta \right| \leq 2\pi r \sup_{\zeta \in \gamma_r} \left|{f(\zeta) \over \zeta - z}\right| \leq 2 \pi {1 \over |z_0-z| - r} \sup_{\zeta \in \gamma_r} |f(\zeta)| $$ which tends to zero as $r \rightarrow 0$.

⬛️

Corollary (Weaker than linear growth implies analyticity at infinity) If $$ \lim_{z \rightarrow \infty} {f(z) \over z} = 0, $$ then $f$ is analytic at infinity.

From these result we can infer that $$\phi(z) = \sqrt{z-1}\sqrt{z+1}$$ is analytic on $\C \backslash [-1,1]$, and $\phi(z) \sim_{z \rightarrow \infty} z$.

Uniqueness results from inferred singularity

Recall that an important ingredient of complex analysis is Liouville's theorem:

Theorem (Liouville) If $f$ is entire and bounded in ${\mathbb C}$, then $f$ must be constant.

We will see that knowledge of the behaviour of $\phi$ can be used to recover by its behaviour at its singularities at $\infty$ and jump on $[-1,1]$. But before that, we already have the following uniqueness results by combining the above with Liouville's theorem:

1. $\phi(z)$ is the unique function analytic in $\C \backslash [-1,1]$ with weaker than pole singularities at $\pm 1$ satisfying $\phi(z) \sim z$ and $$\phi_+(x) - \phi_-(x) = 2\I \sqrt{1-x^2} \qqfor -1 < x <1.$$
2. $\kappa(z) = {1 \over \sqrt{z-1} \sqrt{z+1}}$ is the unique function analytic in $\bar\C \backslash [-1,1]$ with weaker than pole singularities at $\pm 1$ satisfying $\kappa(\infty) = 0$ and $$ \kappa_+(x) - \kappa_-(x) = {-2\I \over \sqrt{1-x^2}}\qqfor -1 < x <1. $$

Demonstration From the phase plot we see it has a branch cut on $[-1,1]$:

On the branch there is the expected jump:

For $x < -1$ the branch cut is removable: we have continuity and therefore analyticity:

3. $\mu(z) = {\log(z -1) - \log(z+1) \over 2 \pi \I}$ is the unique function analytic in $\C \backslash [-1,1]$ with weaker than pole singularities at $\pm 1$ satisfying $\mu(\infty) = 0$ and $$\mu_+(x) - \mu_-(x) = 1 \qqfor -1 < x < 1.$$

Demonstration Here we see from the phase plot of $mu$ that it has a branch cut on $[-1,1]$:

For $-1 < x < 1$ we have the jump $1$:

For $x < -1$ we see that the branch cuts cancel and we have continuity: