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\newcommand{\id}{\mathop{\rm id}\nolimits} \)
SOME COMPLEX THEOREMS IN SIMPLE ANALYSIS
ISHAN LEVY
1. Holomorphic Functions
Let’s recall the statements of two important theorems that we will need. Throughout the first
part of the notes (about holomorphic functions), we assume that \(f\) is a holomorphic
function on an open set \(\Omega \), \(D\) is an open disk. \(D(z_0,r)\) indicates that this disk is centered at \(z_0\) of radius
r.
Theorem 1.1 (Cauchy’s Theorem in a Disk). If \(D\subset \Omega \), then \[ \int _\gamma f(z)dz=0 \] for any closed curve \(\gamma \) in \(D\).
Theorem 1.2 (Cauchy’s Integral Formulæ). \(f\) has infinitely many derivatives on \(\Omega \) and for
any circle \(C \subset \Omega \) that is boundary of \(D \subset \Omega \) we have \[ f^{(n)}(z) = \frac{n!}{2\pi i}\int _C\frac{f(\zeta )}{(\zeta -z)^{n+1}}d\zeta \] for any \(z\) in the interior of \(C\).
As a consequence we have the following:
Corollary 1.3 (Cauchy Inequalities). If \(\Omega \) contains the closure of \(D(z_0,R)\), then \[ |f^{(n)}(z)| \leq \frac{n!|f|_C}{2\pi R^n} \] where \(|f|_C\) is the maximum
modulus \(f\) attains on \(C\).
Proof. By Theorem 1.2 we have: \begin{align*} |f^{(n)}(z)| &= |\frac{n!}{2\pi i}\int _C\frac{f(\zeta )}{(\zeta -z)^{n+1}}d\zeta |\\ &= \frac{n!}{2\pi }|\int _0^{2\pi }\frac{f(z_0 + Re^{i\theta }) Rie^{i \theta }}{(Re^{i\theta )^{n+1}}}d\theta |\\ &\leq \frac{n!}{2\pi }\int _0^{2\pi }|\frac{f(z_0 + Re^{i\theta }) Rie^{i \theta }}{(Re^{i\theta )^{n+1}}}|d\theta \\ &= \frac{n!|f|_CR(2\pi )}{2\pi R^{n+1}}\\ &= \frac{n!|f|_C}{R^{n}} \end{align*}
□
Here is another corollary of the Cauchy Integral Formulæ:
Corollary 1.4. Holomorphic functions are analytic.
Proof. Suppose \(f\) is holomorphic near \(z_0\). By Theorem 1.2 we have \[ f(z) = \frac{1}{2\pi i}\int _C\frac{f(\zeta )}{\zeta -z}d\zeta \] where \(C\) is a small circle
centered around \(z_0\), and \(z\) is inside the disk. Additionally \[ \frac{1}{\zeta -z} = \frac{1}{\zeta -z_0}\frac{1}{1-\frac{z-z_0}{\zeta - z_0}} \] and since \(z\) is inside the circle and
\(\zeta \) is on the boundary, so we can expand as \[ \frac{1}{1-\frac{z-z_0}{\zeta - z_0}} = \sum _0^\infty \Big (\frac{z-z_0}{\zeta -z_0}\Big )^n \] which converges uniformly for all \(\zeta \in C\). Then from
uniform convergence we switch the sum and integral and get \[ f(z) = \sum _0^\infty \frac{1}{2\pi i} \int _C \frac{f(\zeta )}{(\zeta -z_0)^{n+1}}d\zeta (z-z_0)^n = \sum _0^\infty \frac{1}{n!}f^{(n)}(z_0)(z-z_0)^n \] □
We can use the Cauchy Inequalities to deduce that a non-constant entire (ie. holomorphic on \(\CC \))
function is not bounded.
Corollary 1.5 (Liouville’s Theorem). A bounded entire function \(f\) is constant.
Proof. Fix \(z\) arbitrary. Let \(B\) be the bound on \(f\). Integrate around a large circle centered at \(z\) and
use the Cauchy inequalities to get \(|f'(z)| \leq \frac{B}{R}\). Then letting \(R\) tend to infinity, we get \(f'(z) = 0\) so \(f\) is constant by
Corollary 1.4. □
Theorem 1.6. Suppose \(f\) is holomorphic on a connected open set, and the set of zeros has
a limit point, then \(f = 0\).
Proof. Take a limit point \(z_0\) of the set of zeros of \(f\), and write \(f(z) = \sum _i^\infty a_n(z-z_0)^n\), where \(a_i \neq 0\). Then \[ f(z) = a_i(z-z_0)^i(1+g(z-z_0)) \] where \(g\) is analytic
near \(z_0\) and \(g(z-z_0)\) goes to \(0\) as \(z \to z_0\). Then take a sequence of points \(\neq z_0\) converging to \(z_0\). But by the formula
above, some of these points cannot be \(0\) as \(1+g(z-z_0)\) is nonzero near \(z_0\). Then \(f\) is \(0\) near \(z_0\). This shows that
the set of limit points of the zeros of \(f\) is open. But it’s also closed and \(f\) is holomorphic on a
connected set. □
Corollary 1.7 (Analytic continuation). If there are two functions \(f,g\) holomorphic on a
connected open set such that they agree on a set with a limit point, then they are the
same.
Proof. Theorem 1.6 on \(f-g\). □
Analytic continuation is an important principle used to extend functions to larger parts of \(\CC \)
by defining it over a small region and trying to extend it. One of the ways functions
can be extended is the symmetry principle and in particular the Schwarz reflection
principle. Here is a converse of Goursat’s Theorem that we will use to prove the symmetry
principle.
Theorem 1.8 (Morera’s Theorem). If \(f\) is continuous on \(\Omega \) and for any triangle \(T\) we have \(\int _Tf(z)dz = 0\)
then \(f\) is holomorphic.
Proof. By the proof of Goursat’s Theorem, \(f\) has a primitive, and so the derivative of this
primitive, \(f\), is by Corollary 1.4 also holomorphic. □
Now we will prove the symmetry principle. This is a way of extending a holomorphic
function.
Theorem 1.9 (Symmetry principle). Suppose that \(f^+\) and \(f^-\) are holomorphic functions such
that \(f^+\) is holomorphic on a region of the upper half-plane, and \(f^-\) is holomorphic on the
lower half-plane such that they can be extended continuously to an interval of \(\RR \) such that
they agree on the boundary, then they can be extended to a holomorphic function on the
union.
Proof. For any triangle in the union of the regions, we can split it into a bunch of triangles
in the region where \(f^+\) and \(f^-\) are continuous (and possibly also the boundary). Then we can
approximate the triangles with ones in the interior (where \(f^-\) and \(f^+\) are holomorphic) and by
Goursat’s Theorem, the integral on these triangles is \(0\), so by continuity the integral on the
triangles that do hit the boundary are also \(0\). Then any triangle’s integral is \(0\), so the result
follows from Theorem 1.8. □
As a special case we have the Schwarz reflection principle.
Theorem 1.10 (Schwarz reflection principle). Suppose that \(f\) is holomorphic on some \(\Omega \) in the upper
half-plane, and extends continuously to a real-valued function on an interval on the
boundary, \(I\). Then \(f\) can be extended to a holomorphic function on the region \(\Omega \cup -\Omega \cup I\), where
\(f(\bar{z})=\bar{f(z)}\).
Proof. Use Theorem 1.9. □
2. Meromorphic Functions
Sometimes we don’t want to require such a strong condition as holomorphicity, but would rather
have a function that is ”almost” holomorphic, or meromorphic. I offer two ways to think about
this. One way is that a function can have three types of singularities: removable singularities, poles,
and essential singularities. Removable singularities are as their name suggests removable (ie the
function can be extended to fill in the hole). Essential singularities are wild, we would not
like our notion of meromorphic to allow this. Poles, however, are quite reasonable and
interesting singularities, hence we can allow these. Another way to think about meromorphic
functions is as holomorphic functions, except these no longer land in the complex plane.
Instead, we add a point at infinity to \(\CC \), and call this \(\PP ^1\), the Riemann sphere. This has a
complex analytic structure: it can be thought of as two open disks in \(\CC \) glued together.
Then a meromorphic function is a holomorphic function from an open subset of \(\CC \) to
\(\PP ^1\).
First let’s study singularities. A singularity is a point such that \(f\) is defined around the point but
not on that point. We have a few possibilities:
Definition 2.1. A singularity \(z_0\) is removable if \(f\) is bounded near \(z_0\). A singularity is a pole
of order \(i\) if \((z-z_0)^if\) is bounded near \(z_0\) for some \(i\). A singularity is essential if it is neither of the
other two.
Now we can define meromorphic:
Definition 2.2. \(f\) is meromorphic on an open set \(\Omega \) if there is a set of points \(X \subset \Omega \) without a limit
point such that \(f\) has removable singularities and poles on \(X\) and is holomorphic on \(\Omega -X\).
Theorem 2.3. Removable singularities are removable.
Proof. Suppose \(f\) is bounded near \(z_0\) (WLOG set \(z_0 = 0\)). Then we have \(z^2f\) (extended to be \(0\) at \(z_0\)) is
holomorphic near \(0\) but also at \(0\). Then \(f\) is of the form \[ \frac{a_{-2}}{z^2} + \frac{a_{-1}}{z} + a_0 + \cdots \] \(f\) is bounded near zero so \(a_{-2}\) and \(a_{-1}\) are \(0\). □
We can characterize poles in the way that agrees with the second way of thinking about
meromorphic functions:
Theorem 2.4. A point \(z_0\) is a pole iff \(1/f\) is holomorphic near \(z_0\) where \(1/f(z_0)\) is defined to be \(0\).
Proof. By Theorem 2.3 we have that \(f\) having a pole of order \(n\) at \(z_0\) is equivalent to \(f(z) = (z-z_0)^{-n}h(z)\) for some
function \(h\) holomorphic near \(z_0\) that is nonzero at \(z_0\). Then \(f\) is meromorphic implies \(1/f = (z-z_0)^{n}(1/h)(z)\) is holomorphic.
Conversely if \(1/f\) is holomorphic we can write \(1/f\) in the same way, and invert it to get \(f\) in the form
we want. □