Complex Analysis

Lecture 37

Jordan's Lemma

Lemma: If \(f\) is analytic on \( \left\{ z : \Im z \ge 0\right\} \cap \left\{ z : |z| \gt R_0 \right\}\) satisfies \(\big|\,f(z) \le \frac{M}{R^\beta}\big|\;\) \((*)\;\) (\(M,\beta \gt 0 \) constants) on \[ \Gamma_R=\left\{ Re^{i\theta}: R\gt R_0, 0\le \theta \le \pi \right\}. \] Then \[\ds \lim_{R\ra \infty } \int_{\Gamma_R } e^{i\alpha z} f(z)\, dz = 0,\; \forall \alpha \gt 0.\]

Jordan's Lemma

Proof: On \(\Gamma_R\) we have \(z= Re^{i \theta }\), \(dz = i R e^{i\theta}\)

\(\left| \ds\int_{\Gamma_R } e^{i\alpha z} f(z)\, dz \right|\) \(= R \left| \ds\int_{0}^{\pi} e^{i\alpha Re^{i \theta } } f\left(Re^{i \theta }\right)\, d\theta \right|\) \( \le R \ds \int_0^{\pi} \left| e^{i\alpha Re^{i \theta } } f\left(Re^{i \theta }\right)\, d\theta \right|\)

\(\qquad \qquad \qquad = R \ds \int_0^{\pi} \left| e^{i\alpha \left(R \cos \theta + i R \sin \theta \right) } f\left(Re^{i \theta }\right) \right| d\theta\)

\(\qquad \qquad \qquad = R \ds \int_0^{\pi} e^{-\alpha \,R \sin \theta} \left| \, f\left(Re^{i \theta }\right) \right| d\theta\)

\(\qquad \qquad \qquad \overset{(*)}{\le} \ds \frac{M}{R^{\beta -1}} \int_0^{\pi} e^{-\alpha \,R \sin \theta} d\theta\)

\(\qquad \qquad \qquad = \ds \frac{2M}{R^{\beta -1}} \int_0^{\pi/2} e^{-\alpha \,R \sin \theta} d\theta\) \(\quad (**)\)

Jordan's Lemma

Proof (cont): Note that \(\sin \theta \ge \dfrac{2\theta}{\pi}\) for \(0\le \theta \le \pi/2\).

So we have

\((**)\) \(\le \ds \frac{2M}{R^{\beta-1}} \int_0^{\pi/2} e^{-\frac{2\alpha R}{\pi} \theta} d\theta \) \(\qquad\qquad \)

\(= \ds \frac{2M}{R^{\beta-1}} \frac{\pi}{2 \alpha R} \left( 1-e^{-\alpha R} \right) \) \(\qquad \)

\(\quad \quad \;\;= \ds \frac{M \pi }{\alpha R^{\beta}} \left( 1-e^{-\alpha R} \right)\) \(\ra 0\) as \(R\ra \infty\). \(\;\square\)

Integrals \({\small \int_0^{2\pi}f\left(\sin t, \cos t\right)dt}\)

Try the substitution

\(z= \cos t + i \sin t\) \(\;\Rightarrow \; z^{-1} = \cos t + i \sin t\).

Motivation: \(z=e^{it}, 0\le t \le 2 \pi\). \(\Rightarrow \)

\(\cos t = \ds \frac{1}{2} \left(z+z^{-1}\right)\)

\(\sin t = \ds \frac{1}{2i} \left(z-z^{-1}\right)\)

\(\; \Rightarrow \) \(\;dz = (-\sin t + i \cos t) dt = i z dt\).

Integrals \({\small \int_0^{2\pi}f\left(\sin t, \cos t\right)dt}\)

Example: Find \(I = \ds \int_0^{2\pi} \frac{dt}{2+ \cos t} \).

Put \(z = e^{it}\), so by above, \(dt = dz/iz\) and \(\cos t = \frac12 \left(z+z^{-1}\right)\).

So for \(C\) given by \(\left\{ z=e^{it} , 0\le t \le 2 \pi \right\}\), there holds

\(I = \ds \int_C\frac{dz\,/\,iz}{2+\frac{1}{2} \left(z+z^{-1}\right)}\) \(=-i \ds\int_C \frac{dz}{2z + \frac12 \left(z^2+1\right)}\)

\(\;\; =-2i \ds\int_C \frac{dz}{2z^2+4z+1}\) \(\quad (\text{🌏})\).

Integrals \({\small \int_0^{2\pi}f\left(\sin t, \cos t\right)dt}\)

Example (cont): To evaluate the integral in \((\text{🌏})\) note that the integrand is analytic, except at the zeros of the denominator, i. e. \(-2\pm \sqrt{3}\).

Only \(-2 + \sqrt{3}\) is inside the countour \(C\), so by the Residue Theorem and \((\text{🌏})\) we obtain \[I= (-2i)(2\pi i)\underset{{\tiny z=-2+\sqrt{3}}}{\res} \frac{1}{z^2+4z+1} \quad (\text{😀})\]

Integrals \({\small \int_0^{2\pi}f\left(\sin t, \cos t\right)dt}\)

Example (cont): To calculate the residue, note \[\frac{1}{z^2+4z+1} = \frac{\phi(z)}{z - (-2+\sqrt{3})}\] where \[\phi(z) = \frac{1}{z-\left(-2-\sqrt{3}\right)}\] is analytic and non-zero near \(z=-2+\sqrt{3}.\)

Integrals \({\small \int_0^{2\pi}f\left(\sin t, \cos t\right)dt}\)

Example (cont): So \(\dfrac{1}{z^2+4z+1} \) has a pole of order \(1\) at \(z=-2+\sqrt{3}\), and Theorem 2 (Lec 33) implies \[\underset{{\tiny z=-2+\sqrt{3}}}{\res} \frac{1}{z^2+4z+1} = \phi\left(-2 + \sqrt{3}\right) = \frac{1}{2\sqrt{3}}.\]

Hence,\(\;(\text{😀})\) \(\Rightarrow I = (-2i)(2\pi i)\ds \frac{1}{2\sqrt{3}} \) \(= \ds \frac{2\pi}{\sqrt{3}}\). \(\qquad \square\)

Final remarks on Laurent series

Laurent series for \(\dfrac{1}{\sinh z}\) at \(0\). Note that \(\sinh z = 0\) for \(z= n\pi i\) only with \(n\in \Z\).

So \(\dfrac{1}{\sinh z}\) will have a Laurent series on \(0\lt |z|\lt \pi\). Note

\(\;\;\dfrac{1}{\sinh z} = \dfrac{1}{z+ \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots}\) \(= \dfrac{1}{z}\left( \dfrac{1}{1+ \frac{z^2}{3!} + \frac{z^4}{5!} + \cdots}\right)\) \(\;\;(★)\)

using the formula for the sum of geometric series, for \(|z|\) small,...

Final remarks on Laurent series

... so \(\left| \ds \frac{z^2}{3!} + \frac{z^4}{5!} + \cdots \right|\lt 1\), there holds

\((★)\) \(=\ds \frac{1}{z}\frac{1}{1 + \left( \frac{z^2}{3!} + \frac{z^4}{5!} + \cdots \right) }\)

\(\quad \;\;= \ds \frac{1}{z} \Bigg[ 1 - \left( \frac{z^2}{3!} + \frac{z^4}{5!} + \cdots \right) + \left( \frac{z^2}{3!} + \frac{z^4}{5!} + \cdots \right)^2 - \cdots \Bigg]\)

(via formula for geometric series, as long as \(|z|\) is sufficiently small)

Final remarks on Laurent series

Thus \((★)\) \(= \ds \frac{1}{z} \Bigg[ 1 - \frac{z^2}{3!} + z^4\left( -\frac{1}{5!} + \frac{1}{3!3!} \right) - \cdots \Bigg]\)

\(\qquad \quad \;= \ds \frac{1}{z} -\frac{z}{6} + \frac{7z^3}{360} - \cdots \)

In particular, pole of order \(1\) at \(0\), with \[\underset{{ z=0}}{\res} \frac{1}{\sinh z} = 1. \qquad \square\]

Note that it can also be done via division of power series.