Lecture 22

**An extension**

Let \(f\) be analytic in and on \(C\), a simple closed curve in \(\C\), traversed in the positive sense, and let \(z_0\in \text{Int}\,C\). From Lecture 21:

Cauchy \(\Rightarrow f(z_0) = \ds \frac{1}{2\pi i} \int_C \frac{f(z)}{z-z_0} dz\).

**An extension**

**Theorem:**
\[
f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_C \frac{f(z)}{(z-z_0)^{n+1}} dz, \quad (1)
\]
for \(n>1\).

**An extension**

For \(f= u + i v \), \(\,f\) analytic at \(z_0 = x_0+ i y_0\) implies partials of all orders of \(u\) and \(v\) exist and are continuous at \((x_0, y_0)\).

**c. f.** the situation in
\(\R\),
**e. g.** \(f(x)=|x|^3\),
\(f, f' , f''\) are all contiuous on \(\R\), but
\(f'''(0)\) does not exist.

**Note:** If \(f\) is analytic at \(z_0\),
then
its derivatives of all orders exist and are analytic at \(z_0\).

**Proof:** 📝 Check it!

**Theorem:** Let \(f\) be continuous on a domain \(\Omega\).
If
\[
\int_C f = 0,
\]
for every closed contour \(C\) lying in \(\Omega\), then \(f\) is analytic in \(\Omega\).

**Proof:**
From Theorem in
Lecture 19 we know that \(f\) has a primitive on \(\Omega\), say \(F\).

But then \(F'=f\) exists and is continuous on \(\Omega\) by the conditions of this theorem, \(F\) is analytic.

Hence by the previous Note \(\,f=F'\) is also analytic. \(\square\)

**of the Cauchy Integral Formula**

A number of nice results follow from the extension of the Cauchy Integral Formula.

**I.** Let \(f\) be analytic in and
on \(C_R(z_0)\),
and set \(M_R= \ds \max_{z\in C_R}|\,f(z_0)|\).
Then
\[
\left|\,f^{(n)}(z_0)\right| \leq \frac{n!M_R}{R^n} \qquad (2)
\]

**of the Cauchy Integral Formula**

**Proof:**
Note \(M_R\) is well defined (extreme value
theorem).

\(\left|\,f^{(n)}(z_0)\right|\) \(= \left| \ds \frac{n!}{2\pi i} \int_{C_R} \frac{f(z) dz}{(z-z_0)^{n+1}} \right|\) \(\leq \ds \frac{n!}{2 \pi } \int_{C_R} \frac{|\,f(z)|dz}{|z-z_0|^{n+1}}\)

\(\qquad \quad \; \leq \ds \frac{n! M_R}{2 \pi \, R^{n+1} } \int_{C_R} dz\) \(= \ds \frac{n! M_R}{2 \pi \,R \, R^n} 2 \pi \,R\) \(= \ds \frac{n! M_R}{R^n}\).

That is, \(\left|\,f^{(n)}(z_0)\right| \leq\ds \frac{n! M_R}{R^n}\). \(\square\)

**of the Cauchy Integral Formula**

**II. Liouville** If \(f: \C \ra \C\) is bounded and entire. Then \(f\) is
constant.

**of the Cauchy Integral Formula**

**Proof:**
Suppose that \(|\,f|\leq M\) on \(\C\), \(f\)
entire.
Apply \((2)\)
for \(n=1\) on
\[
C_R= \left\{z_0+ Re^{i\theta}, 0\leq \theta \leq 2 \pi \right\}.
\]

Then \[ \left|\,f'(z_0)\right| \leq \frac{1! M}{R} = \frac{M}{R}. \]

Letting \(R\ra \infty\), we see that there must hold: \(f'(z_0) =0\). Since \(z_0\) was arbitrary, we're done. \(\square\)

**of the Cauchy Integral Formula**

**II. Fundamental Theorem of Algebra**