Complex Analysis

Lecture 29

Poisson kernel

Taking the real part from 😀 defined in Lecture 28: Given "nice enough" \(\Phi(r_0, \phi)\) defined on the boundary \(C_0\) of \(B_{r_0}\), a (infact, the) solution of the Dirichlet problem \[ \text{(D) } \left\{ \begin{align} \Delta u = & \;0 \;\text{ in } \;B_{r_0}\\ u|_{\partial B_{r_0}} = & \;\Phi(r_0, \phi) \end{align} \right. \] is given by \(u(r, \theta) = \ds \frac{1}{2 \pi } \int_{0}^{2 \pi} \frac{r_0^2-r^2}{r_0^2-2r_0r \cos \left(\phi - \theta\right) +r^2} \Phi(r, \phi)d\Phi\) \[= \ds \frac{1}{2 \pi } \int_{0}^{2 \pi} P\left(r_0, r, \phi- \theta\right) \Phi(r, \phi)d\Phi \]

where \(P(r_0, r, \phi- \theta)\) is known as the Poisson kernel.


Compare the situation in \(\R\).

Formally, a sequence is a function:

\(\N \ra \C \) (or \(\N_0 \ra \C \)) \[n \mapsto z_n\]

Write: \(\{z_n\}\).


Limit of a sequence:

\(\ds \lim_{n\ra \infty} z_n = z \iff \{z_n\} \) converges to \(z\).

\(\iff\) Given \(\varepsilon >0 \) there exists \(N\in \N\) such that \[n\gt N \implies |z_n-z|\lt \varepsilon.\]


Now for series: Formally we have \(\ds \sum_{n=0}^{\infty} z_n\), \(\,z_n \in \C\).

The series \(\sum z_n\) converges as a series \(\iff\) the associated sequence of partial sums \(\{s_n\}\) converges as a sequence.

Here: \(s_n = \ds \sum_{k=0}^{n} z_k\).


Typical question: Does \(\sum z_n\) converge?

Test for NO: The \(n\)th term test \[\sum z_n \text{ converges } \Rightarrow z_n \ra 0.\]

\(\nLeftarrow\) e. g. \(\sum 1/n\) diverges.


Remark: \(\{z_n\}\) is bounded says there exists \(M\) such tat \(|z_n|\lt M\) for all \(n\).

\(*\) convergent \(\Rightarrow\) bounded, \(\nLeftarrow \{(-1)^n\}\).

Definition: \(\sum z_n\) converges absolutely \(\iff\) \(\sum |z_n|\) converges.

Absolute convergence \(\Rightarrow\) convergence,

\(\nLeftarrow\) \(\sum \frac{(-1)^n}{n}\)


Given \(\sum_{n=0}^{\infty} z_n\), set \(S_n = \sum_{k=0}^{n} z_k\) as above, and write \[\rho_n = \sum_{k=n+1}^{\infty} z_k\]

\(\rho\) is the tail/remainder. Then \[S_n \ra S \iff \sum z_n = S \;\text{ by definition.}\]

Theorem: \(S_n \ra S \iff \rho_n =0\).



Claim \(\ds\sum_{n=0}^{\infty} z^n = \frac{1}{1-z}\) for \(|z|\lt 1\). \(\quad(*)\)

Proof: \(\qquad S_n = 1+ z+ \cdots + z^n\)

\(\qquad \qquad z S_n = z+ \cdots + z^n + z^{n+1}\).

\(\Rightarrow (1-z) S_n = 1-z^{n+1}\) \(\Rightarrow S_n = \dfrac{1-z^{n+1}}{1-z}\quad\qquad\)

\(\Rightarrow \rho_n = S - S_n \) \(=\dfrac{z^{n+1}}{1-z}\). Since \(|z|\lt 1\), \(|\rho_n|\ra 0\) as \(n\ra \infty \),

\(\Rightarrow \rho_n\ra 0\) as \(n\ra \infty\), \(\Rightarrow S_n \ra S\), which shows \((*)\).


Remark: As in \(\R\), if \(\sum a_n\) and \(\sum b_n\) converge, then so do \[\sum \left(a_n\pm b_n\right) \text{ to } \left(\sum a_n\right) \pm \left(\sum b_n\right).\]

For \(\lambda \in \C \) fixed

\(\ds\sum \left( \lambda a_n\right) \) converges to \(\lambda \ds\sum a_n\).

Power Series

Centred at \(z_0\), the series of a function \(f(z)\) is \(\ds\sum_{n=0}^{\infty} a_n (z-z_0)^n.\)

Series will

  • converge absolutely for \(|z-z_0|\lt R\), where \(R\) is the radius of convergence;
  • diverge for \(|z-z_0|\gt R\);
  • check by hand for \(|z-z_0|= R\);
  • \(R=0\), converge only at \(z_0\);
  • \(R=\infty\), converge on \(\C\).