# MATH3401

Lecture 22

### Cauchy Integral formula

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

### Cauchy Integral formula

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 $$n>1$$.

### Cauchy Integral formula

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!

### Morera's Theorem

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 $$\Omega$$.

### Morera's Theorem

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. $$\square$$

### Consequences of the extension

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 $\left|\,f^{(n)}(z_0)\right| \leq \frac{n!M_R}{R^n} \qquad (2)$

### Consequences of the extension

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

### Consequences of the extension

of the Cauchy Integral Formula

II. Liouville   If $$f: \C \ra \C$$ is bounded and entire. Then $$f$$ is constant.

### Consequences of the extension

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. $$\square$$

### Consequences of the extension

of the Cauchy Integral Formula

II. Fundamental Theorem of Algebra