Vrftsjtryk

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 $-3pi /2 < arg{(z pm i)} le pi/2$.

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 $pm z$?

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

$$ left( 1 + z^2right)^{1/2} = z + frac{1}{2z} + Oleft( frac{1}{z^3} right) $$
as $z to infty$.