I'm working on a fairly simple question asking to work out the necessary branch cut(s) for the function $f(z)=(z^2+1)^{1/2}$. I am comfortable doing this and the rigour required to explained why I need / don't need certain branch cuts.
I have a branch cut between $(-i, i)$ on the imaginary axis and my argument is defined as $-3\pi /2 < \arg{(z \pm i)} \le \pi/2$.
The next part of the questions says
For your branch, show also that as $|z|\rightarrow \infty $ we have,
$$f(z)=z+\frac{1}{2z} + O(\frac{1}{z^3}).$$
In a comparable example, this is shown by computing the binomial expansion for the function and then leaping to saying that this is equal to the Laurent Expansion.
Here's my working:
$f(z)$ is regular everywhere except at $z= \pm i$ and our branch cut. Hence, it has a Laurent series. On the +ve real axis we have $f(z)=(x^2+1)^{1/2}=|x|(1+(1/x)^2)^{1/2}$ which I can expand by using the binomial theorem like so,
$$f(z)=|x|(1+\frac{1}{2x^2}-\frac{1}{8x^4}+… \quad=x+\frac{1}{2x}+O(\frac{1}{x^3}).$$
I believe all that is left to do is say that this is true for $|z|\rightarrow \infty$, but how can I?
My main questions are:
For what values of $x$ is the binomial expansion valid?
Why can I suddenly say that this expansion characterises the behaviour of $f(z)$ for large $\pm z$?
Best Answer
The key point is that $(1 + w)^{1/2}$ has a Taylor series expansion around $w = 0$, given by $$ (1 + w)^{1/2} = 1 + \frac {w}{2} - \frac {w^2}{8} + \frac{w^3}{16} - \dots$$ Since $(1 + w)^{1/2}$ is holomorphic on the unit disk $B(0, 1)$, this series expansion is valid on the whole of the disk $B(0, 1)$.
[This follows from the standard proof of Taylor's theorem for holomorphic functions. The general result is that if $f : U \to \mathbb C$ is holomorphic and $\bar B(w_0, r) \subset U$, then the Taylor series for $f$ around $w_0$ is uniformly convergent and equal to $f$ on $B(w_0, r)$. See here for the proof. Applying this to $f(w) = (1 + w)^{1/2}$ with $U = B(0, 1)$ and $w_0 = 0$, we find that the above series expansion is uniformly convergent and equal to $(1 + w)^{1/2}$ on $B(0, r)$, for every $r < 1$. Hence it is also pointwise convergent and equal to $(1 + w)^{1/2}$ on $B(0, 1)$ itself.]
[I should also clarify, in case it isn't obvious, that we're considering the branch of $(1 + w)^{1/2}$ that takes the value $1$ when $w = 0$. This branch is well defined on the whole of the unit disk $B(0, 1)$.]
Anyway, if we now substitute $w = z^{-2}$, we see that the series expansion $$ \left( 1 + z^{-2}\right)^{1/2} = 1 + \frac{1}{2z^2} - \frac{1}{8z^4} + \frac{1}{16 z^6} - \dots$$
is valid for all $z$ such that $|z| > 1$. So this series expansion certainly captures the behaviour of $\left( 1 + z^{-2} \right)^{1/2}$ as $z \to \infty$.
I think it might help to spell out how to rigorously arrive at the result about the asymptotic behaviour. The big-O notation is a bit confusing here.
So the Taylor series expansion for $(1 + w)^{1/2}$ tells you that there exists a holomorphic function $g : B(0,1) \to \mathbb C$ such that $$ (1 + w)^{1/2} = 1 + \frac{w}{2}+ w^2 g(w)$$ for all $w \in B(0, 1)$.
By continuity, there certainly exists an $M$ and a $\delta > 0$ such that $$|w| < \delta \implies g(w) \leq M$$ i.e. such that $$|w| < \delta \implies (1 + w)^{1/2} - 1 - \frac{w}{2} \leq Mw^2.$$
Setting $w = z^{-2}$, and multiply through by $z$ we find that $$ |z| > \frac 1 {\sqrt{\delta}} \implies \left( 1 + z^2 \right)^{1/2} - z - \frac{1}{2z} \leq \frac{M}{z^3},$$ which is to say that
$$ \left( 1 + z^2\right)^{1/2} = z + \frac{1}{2z} + O\left( \frac{1}{z^3} \right) $$ as $z \to \infty$.