Proving that two branch cuts can cancel out

Question

Define the following functions $\mathbb{C}\to\mathbb{C}:$ $$u(z)=\frac{\log \left(z+\frac{1}{2}\right)}{z}\quad \left[-\pi\leqslant\arg \left(z+\tfrac12\right)<\pi\right];\quad v(z)=\frac{\log z}{z}\quad \left[-\pi\leqslant\arg z<\pi\right]$$ The function $f=u-v$ is holomorphic on $1<|z|<R$ for any $R>1$. The way I picture this is that the arguments essentially "cancel out" when we cross $(-\infty,-1)$, which lies within the intersection of the branch cuts $(-\infty,0]$ and $(-\infty,-\frac12]$:

• How would I argue this formally though? If write it out we have: $$\frac{1}{z}\log\left|1+\frac{1}{2z}\right|+\frac{1}{z}\left(\arg \left(z+\frac12\right)-\arg z\right)$$ But I can't see exactly why the arguments cancel?

I am asked to find the first terms of the Laurent expansion about $0$ in an annulus as above, and deduce the integral of $f$ around a circle of radius $1<r<R$ centered at $0$. I get: $$f(z)=\frac{1}{z}\log\left(1+\frac{1}{2z}\right)=\frac{1}{2z}\left(\frac{1}{2z}-\frac{1}{8z^2}+\frac{1}{24z^3}-\cdots\right)$$ which vanishes? The way the question is worded makes me think I should be getting something more interesting than just $0$, have I gone wrong?

Answer 1

First, $u$ and $v$ are not defined at $-\frac 12$ and $0$ respectively so you can't pretend they are functions $\Bbb C \to \Bbb C$.

From the properties of the principal branch of $\log$, $u$ and $v$ are holomorphic on $U = \Bbb C \setminus \Bbb R_{\le 0}$. Their difference $u-v$ is also holomorphic on $U$.

Next, $\log(1+\frac 1{2z})$ is holomorphic on $V = \Bbb C \setminus [-\frac 12 ; 0]$.

Since $U \subset V$, we have a holomorphic function $h$ defined on $U$ by $h(z) = \log(z+\frac 12) - \log(z) - \log(1+\frac 1{2z}) \in 2ik\pi\Bbb Z$.
Since $U$ is connected and $\Bbb Z$ is discrete, $h$ has to be constant on $U$. Looking at what happens at $z=1$ we have $h(1) = \log \frac 32 - \log 1 - \log \frac 32 = 0$, and so $h(z) = 0$ forall $z \in U$.

This shows that $\log(z+\frac 12) - \log(z)$, even though we only know (from its expression) that it is holomorphic on $U$, it coincides on $U$ with $\log(1+\frac 1{2z})$, which is holomorphic on $V$.
Since the values of $\log(z+\frac 12) - \log(z)$ on $(-\infty ; -2)$ are the limit of the values of the expression when you approach the real axis from below, they also coincide with $\log(1+\frac 1{2z})$ on $(-\infty ; -2)$ because it is continuous there.

And so $ \forall z \in V, u(z)-v(z) = \frac 1z \log(1 + \frac 1{2z})$, which is holomorphic on $V$ .

Letting $w = \frac 1z$, we have $u(z)-v(z) = w\log(1+w/2)$, whose power series at $w=0$ has radius of convergence $2$ : $w\log(1+w/2) = w^2/2 - w^3/8 + w^4/24 - w^5/64 \dots$. The coefficient of $z^{-1} = w$ is $0$ so your integral is $0$.

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Another way to see that the integral is $0$ is to compare the integrals of $u$ and $v$ on a large circle.

Since (as $|z| \to \infty$), $\log(z+\frac 12) = \log(z) + O(1/|z|)$, you have $u(z) = v(z) + O(1/|z|^2)$. And so the difference of their integral on a circle of radius $R$ is a $O(1/R)$. Since it doesn't depend on $R$, this difference has to be $0$.

Answer 2

We can look at the power series for the difference: $$ \begin{align} u-v &=\frac{\log\left(z+\frac12\right)}z-\frac{\log(z)}z\\ &=\frac{\log\left(1+\frac1{2z}\right)}z\\ &=\frac1{2z^2}-\frac1{8z^3}+\frac1{24z^4}-\frac1{64z^5}+\cdots+\frac{(-1)^{n+1}}{n2^nz^{n+1}}+\cdots \end{align} $$ This series converges for $\left|z\right|\gt\frac12$ and has $0$ residue (the coefficient of the $\frac1z$ term is $0$).

Two branches may differ by a constant, so if we evaluate at $z=1$, we get that the difference of these functions is $$ \log\left(\frac32\right)=\sum_{k=1}^\infty\frac{(-1)^{n+1}}{n2^n} $$ which agrees with the formula above.