Mathematical Methods for Physics and Engineering - Matematica.NET

COMPLEX VARIABLESwhere a is a finite, non-zero complex number. We note that if the above limit isequal to zero, then z = z 0 is a pole of order less than n, orf(z) is analytic there;if the limit is infinite then the pole is of an order greater than n. It may also beshown that if f(z) has a pole at z = z 0 ,then|f(z)| →∞as z → z 0 from anydirection in the Argand diagram. § If no finite value of n can be found such that(24.24) is satisfied, then z = z 0 is called an essential singularity.◮Find the singularities of the functions(i) f(z) = 11 − z − 1 , (ii) f(z) =tanhz.1+z(i) If we write f(z) asf(z) = 11 − z − 11+z = 2z(1 − z)(1 + z) ,we see immediately from either (24.23) or (24.24) that f(z) has poles of order 1 (or simplepoles) atz =1andz = −1.(ii) In this case we writef(z) =tanhz = sinh z exp z − exp(−z)=cosh z exp z +exp(−z) .Thus f(z) has a singularity when exp z = − exp(−z) or, equivalently, whenexp z =exp[i(2n +1)π]exp(−z),where n is any integer. Equating the arguments of the exponentials we find z =(n + 1 )πi, 2for integer n.Furthermore, using l’Hôpital’s rule (see chapter 4) we havelimz→(n+ 1 2 )πi{[z − (n +12 )πi]sinhzcosh z}= limz→(n+ 1 2 )πi{[z − (n +12Therefore, from (24.24), each singularity is a simple pole. ◭})πi]coshz +sinhz=1.sinh zAnother type of singularity exists at points for which the value of f(z) takesan indeterminate form such as 0/0 but lim z→z0 f(z) exists and is independentof the direction from which z 0 is approached. Such points are called removablesingularities.◮Show that f(z) =(sinz)/z has a removable singularity at z =0.It is clear that f(z) takes the indeterminate form 0/0 atz = 0. However, by expandingsin z as a power series in z, we findf(z) = 1 (z ···)− z3z 3! + z55! − =1− z23! + z45! − ··· .§ Although perhaps intuitively obvious, this result really requires formal demonstration by analysis.838