Complex Analysis

Math 312  MWF 10:30am - 11:25am
\begin{matrix}c \\ ◯ \end{matrix}1998 Ron Buckmire  Fowler 112

Class 29 (Monday April 6)

SUMMARY Classifying Singularities and Representing Functions using Series CURRENT READING Brown & Curchill pages 150-154 NEXT READING Brown & Curchill pages 138-156

Cauchy’s Second Residue Theorem

If a function f(z) is analytic everywhere in the �nite plane except for a �nite number of singular points interior to a positively oriented simple closed contour C, then

∮ f(z) dz = 2πiRes [\frac{1}{z^{2}}f (\frac{1}{z}); 0]

In other words, instead of �nding the residues of all the singularities of the given function f(z) which lie inside the given contour C, all you need to do is �nd the residue at a single

point, z = 0, of the associated function \frac{1}{z^{2}}f(\frac{1}{z}). Note what’s really going on involves �nding the residue of the function at the point at in�nity. Exercise Evaluate ∮_{|z|=2π} tan(z) dz using Cauchy’s Second Residue Theorem.

Groupwork Evaluate ∮_{|z|=2} \frac{3z + 2}{z^{2} + 1} dz using Cauchy’s Second Residue Theorem.

Classifying Singularities

There are basically three types of singularities (points where f(z) is not analytic) in the complex plane. They are called removable singularities, isolated singularities and branch singularities.

Isolated Singularity An isolated singularity of a function f(z) is a point z_{0} such that f(z) is analytic on the punctured disc 0 < |z − z_{0}| < r but is unde�ned at z = z_{0}. We usually call isolated singularities poles. An example is z = i for the function z/(z − i).

Removable Singularity A removable singularity is a point z_{0} where the function f(z_{0}) appears to be unde�ned but if we assign f(z_{0}) the value w_{0} with the knowledge that lim_{z→z_{0}} f(z) = w_{0} then we can say that we have �removed� the singularity. An example would be the point z = 0 for f(z) = sin(z)/z.

Branch Singularity A branch singularity is a point z_{0} through which all possible branch cuts of a multi-valued function can be drawn to produce a single-valued function. An example of such a point would be the point z = 0 for Log (z).

There is also a special kind of isolated singularity, called an essential singularity. The canonical example of an essential singularity is z = 0 for the function f(z) = e^{1/z}. The easiest way to de�ne an essential singularity of a function involves Laurent Series (see below). Laurent series In fact, the best way to identify an essential singularity z_{0} of a function f(z) (and an alternative way to compute residues) is to look at the series representation of the function f(z) about the point z_{0} That is,

f(z) = ∑_{n=0}^{∞}a_{n}(z − z_{0})^{n} + ∑_{n=1}^{∞} \frac{b_{n}}{(z − z_{0})^{n}} , R_{1} < |z − z_{0}| < R_{2}

This formula for a Laurent series is sometimes written as

f(z) = ∑_{n=−∞}^{∞} c_{n}(z − z_{0})^{n} where c_{n} = \frac{1}{2πi} ∮_{C} \frac{f(z)}{(z − z_{0})^{n+1}}dz, n = ±1, ±2, …

This �rst part of this series should look somewhat familiar from your experience with real functions, since the expression is clearly a Taylor series if b_{n} = 0 for all n. This �rst part of the series representation is known as the analytic part of the function. The second part (with the negative exponents) is called the principal part of the function. However if a_{n} and b_{n} are not all identically zero this type of series is called a Laurent series and converges to the function f(z) in the annular region R_{1} < |z − z_{0}| < R_{2}. Exercise Let’s show why expressing the function f(z) in terms of a Laurent Series is useful by proving that the value of the Res(f; z_{0}) is exactly equal to b_{−1} (or c_{−1}), that is, the coef�cient

of the \frac{1}{z − z_{0}} term. We can do this by integrating the Laurent series term by term on some

closed contour C and using the CIF.

Review of Sequences and Series Recall that an in�nite sequence {z_{n}} converges to z if for each ϵ > 0 there exists an N such that if n > N then |z_{n} − z| < ϵ The sequence z_{1}, z_{2}, z_{3}, …, z_{n}, … converges to the value z = x + iy if and only if the sequence x_{1}, x_{2}, x_{3}, … converges to x and y_{1}, y_{2}, y_{3}, … converges to y. In other words lim_{n→∞} z_{n} = z ⇔ lim_{n→∞} x_{n} = x and lim_{n→∞} y_{n} = y

An in�nite series ∑_{n=1}^{∞}z_{n} = z_{1} + z_{2} + z_{3} + ⋯ + z_{n} + ⋯ converges to S if the sequence S_{N}

of partial sums where S_{N} = z_{1} + z_{2} + z_{3} + z_{4} + ⋯ + z_{N} (N = 1, 2, 3, …) converges to

S. Then we say that ∑_{n=1}^{∞}z_{n} = S.

As with sequences, series can be split up into real and imaginary parts. Suppose

z_{n} = x_{n} + iy_{n} and ∑_{n=1}^{∞}z_{n} = Z, ∑_{n=1}^{∞}x_{n} = X and Y = ∑_{n=1}^{∞}y_{n} then Z = X + iY.

Taylor series Suppose a function f is analytic throughout an open disk |z − z_{0}| < R_{0} centered at z_{0} with radius R_{0}. Then at each point z in this disk f(z) has the series representation

f(z) = ∑_{n=0}^{∞}a_{n}(z − z_{0})^{n} where a_{n} = \frac{f^{(n)}(z_{0})}{n!} for (n = 0, 1, 2, …)

In other words the function f(z) can be represented exactly by the in�nite series in the disk |z − z_{0}| < R When z_{0} = 0 the series is known as a Maclaurin series. Here are some examples of well known Maclaurin series you should know.

e^{z}  =  1 + z + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + …  = ∑_{k=0}^{∞} \frac{z^{k}}{k!}  |z| < ∞
sin(z)  =  z − \frac{z^{3}}{3!} + \frac{z^{5}}{5!} − …  = ∑_{k=0}^{∞} \frac{(−1)^{k+1}z^{2k+1}}{(2k + 1)!}  |z| < ∞
cos(z) =  1 − \frac{z^{2}}{2!} + \frac{z^{4}}{4!} − …  = ∑_{k=0}^{∞} \frac{(−1)^{k+1}z^{2k}}{(2k)!}  |z| < ∞
sinh(z) =  z + \frac{z^{3}}{3!} + \frac{z^{5}}{5!} + …  = ∑_{k=0}^{∞} \frac{z^{2k+1}}{(2k + 1)!}  |z| < ∞
cosh(z) =  1 + \frac{z^{2}}{2!} + \frac{z^{4}}{4!} + …  = ∑_{k=0}^{∞} \frac{z^{2k}}{(2k)!}  |z| < ∞
\frac{1}{1 − z}  =  1 + z + z^{2} + z^{3} + …  = ∑_{k=0}^{∞}z^{k}  |z| < 1
ln(1 + z) =  z − \frac{z^{2}}{2} + \frac{z^{3}}{3} + …  = ∑_{k=1}^{∞} \frac{z^{k}}{k}  |z| < 1
(1 + z)^{p} = 1 + pz + \frac{p(p − 1)z^{2}}{2!} + … \frac{p(p − 1) … (p − n + 1)}{n!} …  |z| < 1
tan(z)  =  z + \frac{z^{3}}{3} + \frac{2z^{5}}{15} + …  |z| < \frac{pi}{2}

\underline{Groupwork}

1. Write down the MacLaurin series for f(z) = e^{1/z}.

2. What is the value of Res(e^{1/z}, 0)?

3. Evaluate ∮_{|z|=1} e^{1/z}dz two different ways.