Problem 35 - A First Course in Complex Analysis with Applications

Question

(a) Use series to show that \(z=0\) is a zero of order 2 of \(1-\cos z\). (b) In view of part (a), \(z=0\) is a pole of order two of the function \(f(z)=e^{z} /(1-\cos z)\) and hence has a Laurent series $$ f(z)=\frac{e^{z}}{1-\cos z}=\frac{a_{-2}}{z^{2}}+\frac{a_{-1}}{z}+a_{0}+a_{1} z+a_{2} z^{2}+\cdots $$ valid for \(0<|z|<2 \pi\). Use series for \(e^{z}\) and \(1-\cos z\) and equate coefficients in the product $$ e^{z}=(1-\cos z)\left(\frac{a_{-2}}{z^{2}}+\frac{a_{-1}}{z}+a_{0}+\cdots\right) $$ to determine \(a_{-2}, a_{-1}\), and \(a_{0}\). (c) Evaluate \(\oint_{C} \frac{e^{z}}{1-\cos z} d z\), where \(C\) is \(|z|=1\).

Step-by-Step Solution

Answer

(a) Zero of order 2; (b) \(a_{-2}=2\), \(a_{-1}=0\), \(a_0=1\); (c) Integral is 0.

Step 1: Expand the series for cos z

The Taylor series for \( \cos z \) around \( z = 0 \) is given by \( \cos z = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots \). This means \( 1 - \cos z = 1 - \left(1 - \frac{z^2}{2!} + \cdots\right) = \frac{z^2}{2!} - \frac{z^4}{4!} + \cdots \).

Step 2: Identify the order of zero at z = 0 for 1 - cos z

From the series expansion, observe that the lowest power of \( z \) is \( z^2 \). Thus, \( z = 0 \) is a zero of order 2 for \( 1 - \cos z \).

Step 3: Expand the series for e^z

The Taylor series for \( e^{z} \) around \( z = 0 \) is \( e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \).

Step 4: Formulate product series expansion

We assume \( \frac{e^z}{1-\cos z} = \frac{a_{-2}}{z^2} + \frac{a_{-1}}{z} + a_0 + a_1z + \cdots \). The equality \( e^z = (1-\cos z) \left(\frac{a_{-2}}{z^2} + \cdots \right) \) is expanded as \((\frac{z^2}{2} - \cdots)(\frac{a_{-2}}{z^2} + \frac{a_{-1}}{z} + a_0 + a_1z + \cdots)\).

Step 5: Match the coefficients of series

From matching the lowest power terms: for \( z^0 \), the coefficient must match \( 1 \). Thus, \( \frac{z^2}{2} \cdot \frac{a_{-2}}{z^2} = 1 \), giving \( a_{-2} = 2 \). For the \( z^1 \) term: \( \frac{z^2}{2} \cdot \frac{a_{-1}}{z} = z \) should equal 0, so \( a_{-1} = 0 \). For \( z^2 \): \( \frac{z^2}{2} \cdot a_0 = \frac{z^2}{2} \) gives \( a_0 = 1 \).

Step 6: Evaluate the integral over contour C

Since \( z=0 \) is a pole of order 2, the residue at \( z=0 \) is \( a_{-1} \), which is 0. Thus, by the residue theorem, \( \oint_C \frac{e^z}{1-\cos z} dz = 0 \).

Key Concepts

Laurent series

A Laurent series is a representation of a complex function as a power series, which can include terms with negative powers. It is particularly useful for understanding functions that have singularities (points where they are not defined or not analytic). These series are expressed as: \[ f(z) = rac{a_{-n}}{z^n} + rac{a_{-n+1}}{z^{n-1}} + rac{a_{-2}}{z^2} + rac{a_{-1}}{z} + a_0 + a_1z + a_2z^2 + rac{a_3}{z^3} + \ \]This series is valid when \( z \) lies within an annular region. Now, let's consider why they're important:

• Negative Powers: These indicate terms that are not present in a Taylor series. They allow capturing the behavior near singularities like poles.
• Dual Nature: They split into two series: the Taylor series part (positive powers) valid within a disk and the principal part (negative powers).

The function \( \frac{e^{z}}{1-\cos z} \) exhibits a Laurent series because it has a pole at \( z=0 \). This pole directs us to analyze the series for terms with negative powers to evaluate integrals or locate residues.

Taylor series

The Taylor series is used to approximate a complex function by an infinite sum of terms calculated using the function's derivatives at a single point. Consider the series for \( f(z) \) centered at the origin:\[ f(z) = f(0) + f'(0)z + \frac{f''(0)}{2!}z^2 + \frac{f'''(0)}{3!}z^3 + \cdots \]This expression is vital for functions that are analytic (derivatives of all orders exist).

• Analytic Functions: Functions everywhere derivatives exist can use Taylor series for evaluations, eliminating complex calculations.
• Convergence: Taylor series converge within a radius around the center point, which can be estimated using the distance to the nearest singularity.

In our exercise, we approximate \( \cos z \) and \( e^{z} \)using Taylor expansions. For \( \cos z \), the series is expanded as:\( 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots \)Effectively, this expansion helps in identifying the presence of zeros and constructing the \( 1 - \cos z \)answer, key for calculating Laurent series.

Residue theorem

The residue theorem is a powerful tool in complex analysis, giving the integral of a complex function around a closed curve in terms of the sum of residues inside the curve. It's stated as:\[ \oint_C f(z) \, dz = 2\pi i \sum \text{Res}(f, z_k) \]where the residue\( \text{Res}(f, z_k) \)is the coefficient\( a_{-1} \)in the Laurent series expansion of\( f \) at pole\( z_k \).

• Simplicity: It reduces difficult contour integrals to simple residue summations.
• Application: Greatly applied in evaluating integrals and finding solutions of functions with complex poles.

For this exercise, since\( z = 0 \)is a pole of order 2 for\( \frac{e^z}{1-\cos z} \),the residue there is computed as\( a_{-1} \),which is 0. Hence, the integral around any closed curve enclosing that pole, like\( |z| = 1 \),vanishes because\( 2\pi i \times 0 = 0 \). This usefulness makes solving complex integrals much more manageable.