7.5 Cauchy Inequality
Theorem 7.9. Let \(f(z)\) be analytic in a simply connected domain \(D\) containing a circular contour \(C\)
of radius \(R\) centred on \(z_0\). If at each point \(z\) on \(C\), \(\,\left |f(z)\right | \leq M\,\), then
\[\left |f^n(z_0)\right | \leq \frac {n! M}{R^n}\,, \quad \text {for}\, n = 1, 2, 3, \cdots \]
Proof. Use the Cauchy integral formula, set \(\, z - z_0 = Re^{i\theta }\, , \quad 0\leq \theta \leq 2\pi \,, \quad dz = iRe^{i\theta }d\theta \) \begin {align*} f^n(z_0) & = \frac {n!}{2\pi i}\int \frac {f(z)}{\big (z - z_0\big )^{n+1}}dz\\\\ & = \frac {n!}{2\pi i}\int ^{2\pi }_0\frac {f\Big (z_0 + Re^{i\theta }\Big )}{\Big (Re^{i\theta }\Big )^{n+1}}\cdot iRe^{i\theta }d\theta \\\\ & = \frac {n!}{2\pi i}\int ^{2\pi }_0\frac {f\big (z_0 + Re^{i\theta }\big )\cdot iRe^{i\theta }}{R^{n + 1} e^{i\theta (n + 1)}} d\theta \end {align*}
Taking modulus \begin {align*} \left |f^n(z_0)\right | & \leq \frac {n!}{2\pi i}\int ^{2\pi }_0\left |\frac {f\big (z_0 + Re^{i\theta }\big )\cdot iRe^{i\theta }}{R^{n + 1} e^{i\theta (n + 1)}}\right |d\theta \\\\ & \leq \frac {n!}{2\pi i}\int ^{2\pi }_0\frac {M}{R^n}d\theta \\\\ & = \frac {n!M}{2\pi R^n}\int ^{2\pi }_0 d\theta \\\\ & = \frac {n!M}{R^n} \end {align*}
Thus \(\quad \left |f^n(z_0)\right | \leq \frac {n!M}{R^n}\) □
Example 7.10. Show that \(\, f(z) = \frac {1}{z + 1}\,\) satisfies the Cauchy inequality on the circle \(\,\left |z - 4\right | = 3\).
Working
\(f(z) = \frac {1}{z + 1}\,\) is analytic everywhere except at \(\, z = -1\). Hence \(f(z)\) is analytic inside the region \(\, \left |z - 4\right | = 3\).
\(f'(z) = \frac {-1}{\big (z + 1\big )^2}\)
\(f''(z) = \frac {2}{\big (z + 1\big )^3}\)
\(f'''(z) = \frac {-6}{\big (z + 1\big )^4}\)
\(f^4(z) = \frac {4!}{\big (z + 1\big )^5}\)
\(\vdots \)
\(f^n(z) = \frac {(-1)^n\,n!}{\big (z + 1\big )^{n + 1}}\)
Thus
\(\left |f^n(z)\right | = \left |\frac {(-1)^n\,n!}{\big (z + 1\big )^{n + 1}}\right | = \frac {n!}{\big (z + 1\big )^{n + 1}}\)
\(\left |f^n(4)\right | = \frac {n!}{\big (4 + 1\big )^{n + 1}} = \frac {n!}{5^{n + 1}}\)
\(\frac {n! M}{R^n} = \frac {n! M}{3^n}\)
\(M\) is the maximum value of \(\left |f(z)\right |\) where \(z\) is on or inside \(\, \left |z - 4\right | = 3\)
\[\therefore \quad \left |f(z)\right | \leq \frac {1}{2}\]
Set \(\, M = \frac {1}{2}\)
\[\therefore \quad \frac {n! M}{R^n} = \frac {n! M}{3^n} = \frac {n!}{2\cdot 3^n}\]
Theorem 7.11. Let \(f(z)\) be analytic for \(\, \left |z\right | < R\,\) and such that \(\, \left |f(z)\right | < M\,\) then \[\left |f^m\big (re^{i\theta }\big )\right | \leq \frac {m! MR}{\big (R - r\big )^{m+1}}\,,\quad \text {for}\, m = 0, 1, \cdots \quad 0\leq r\leq R\]
Proof. From the Cauchy integral formula for higher derivatives \[f^m(z_0) = \frac {m!}{2\pi i}\int _C\frac {f(z)}{\big (z - z_0\big )^{m + 1}}dz\] For a simple closed curve around \(z_0\). Take \(C\) to be the circle \(\,\left |z\right | = R\,\) so that \(\, z = Re^{i\theta },\\ \quad 0\leq \theta \leq 2\pi \). Let \(\, z_0 = re^{i\phi }\,\) with \(\, 0\leq r \leq R\)
\[f^m\big (re^{i\phi }\big ) = \frac {m!}{2\pi i}\int ^{2\pi }_0 \frac {f\big (Re^{i\theta }\big )}{\big (Re^{i\theta } - re^{i\phi }\big )^{m+1}}\,iRe^{i\theta }d\theta \]
\begin {align*} \left |f^m\big (re^{i\phi }\big )\right | & = \left |\frac {m!}{2\pi i}\displaystyle {\int ^{2\pi }_0 }\frac {f\big (Re^{i\theta }\big )}{\big (Re^{i\theta } - re^{i\phi }\big )^{m+1}}\,iRe^{i\theta }d\theta \right |\\\\ & \leq \frac {m!}{2\pi i}\int ^{2\pi }_0\left |\frac {f\big (Re^{i\theta }\big )}{\big (Re^{i\theta } - re^{i\phi }\big )^{m+1}}\,iRe^{i\theta }\right |d\theta \\\\ & \leq \frac {m!}{2\pi }\int ^{2\pi }_0\frac {\left |f\big (Re^{i\theta }\big )\right | \cdot R d\theta }{\big (R - r\big )^{m + 1}}\\\\ & \leq \frac {m! RM}{2\pi \big (R - r\big )^{m + 1}}\int ^{2\pi }_0d\theta \\ & = \frac {m! MR}{\big (R - r\big )^{m + 1}}\\\\ \end {align*} □
Assignment
Let \(f(z)\) be analytic everywhere in the finite complex plane and take \(a\) and \(b\) be any two distinct points
inside the circle \(C\) with equation \(\,\left |z\right | = R.\,\) Using the Cauchy integral formula and considering the difference \(\,f(a) - f(b)\,\)
prove that if \(f(z)\) is bounded, then \(f(z) \equiv \) constant.
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