MA4410 Homework 6 (Due Mar 13)

Spring 2017 - Piret

1
1. Consider the function f (z) = z(z−2)(z 2 +1)
.

(a) Give a closed form expression (as simplified as you can) for the Laurent coefficients when f (z)
is expanded in a form valid near the origin. We need to expand near the origin. Recall that
Taylor series converge from the center of expansion in a disk of of radius R centered at the center
of expansion, where R is the distance from the center of expansion to the nearest singularity.
Furthermore, recall that Laurent Series converge in Annuli from one singularity to the next. To
see what is going on better, let use use partial fraction decomposition to rewrite f (z) as the
sum of three terms:
−1 1 2z − 1
f (z) = + +
2z
|{z} 10(z − 2) 5(1 + z 2 )
| {z } | {z }
(1) (2) (3)

Before proceeding, we note that there are many poles to this function. (1) has a pole at the
origin. (2) has a pole at z = 2. (3) has two poles, one at z = i and the other at z = i. Thus,
expanding at the origin, (1) would have to be a Laurent expansion, valid in an annulus from
the origin to infinity, (2) would have a Taylor expansion valid in a disk of radius 2 centered at
the origin, and (3) would have a Taylor expansion valid in a circle of radius 1. In conclusion,
for part (a), we are seeking a Laurent expansion for f (z) (Taylor Series being special cases of
Laurent Series) valid up to i. Let us find the Laurent Series: (1) is already a Laurent Series.
(2) can beP∞rewritten as a Geometric PSeries (Taylor) as it is valid in our region of interest.
∞
(2) = −1
20 k=0 (z/2) k
. And (3) = 2z−1
5
k 2k
k=0 (−1) (z) . Totaling, we arrive at

∞
1 X
f (z) = − − Ck z k
2z k=0

where
−1
Ck = − 1/5, k even
2k 20
−1
= k − 2/5, k odd
2 20

This series will be valid in the annulus 0 < |z| < i. Below is a plot in Mathematica of the
Laurent Series for this part of the question. Notice that the series converges in the annulus
0 < |z| < i, and diverges outside of it.
(b) Give the coefficients c−1 , c0 , c1 in a Laurent expansion for f (z) around the origin, valid far out
from the origin. Now we need to expand at the origin for a Laurent Series valid far out from the
origin. That is, in an annulus from 2 to ∞. In this region, (1) will still converge, however, (2)
 Figure 1: Magnitude of f(z) and then of the Laurent Series

and (3) written out as Geometric Series will not converge. We now need to write down Laurent
Series for both (2) and (3). To do this, notice that:
  ∞
1 1 1 1 X
= = (2/z)k
10(z − 2) 10z 1 − 2/z 10z k=0
| {z }
Geometric

The Geometric Series will converge outside of the disk of radius 2 in which (2) originally con-
verged. Also notice that:
  ∞
2z − 1 2z − 1 1 2z − 1 X
2
= 2 2
= 2
(−1)k (1/z 2 )k
5(1 + z ) 5z 1+z 5z k=0
| {z }
Geometric

The series here will converge outside of a disk of radius 1, which overlaps with the region we
seek to expand in. Totaling, we arrive at:
∞ ∞
1 1 X k 2z − 1 X
f (z) = − + (2/z) + 2
(−1)k (1/z 2 )k
2z 10z k=0 5z k=0
 Please note that the above Series is not written out in standardPform. You would have to
multiply the last term out and collect terms to put it in the form Ck z k . This Laurent Series
will converge for 2 < |z| < ∞. In this case, the coefficients c−1 = c0 = c1 = 0. Below are some
plots of the function and of the Laurent Series far out from the origin:

Figure 2: Magnitude of f(z) and then of the Laurent Series

(c) Give the similar three coefficients c−1 , c0 , c1 for a Laurent expansion valid in the vicinity of
z = i. For the Laurent expansion valid in the vicinity of z = i we can shift f (z) down one unit
and then expand at the origin. To do so, we define a new variable t = z?i. The function then
becomes
i 1 1/5 + i/10 1/5 − i/10
f (t) = − + +
2(1 − it) 10(i − 2 + t) | {z t } | 2i{z
+t }
| {z } | {z }
(1c) (2c) (3c) 4c)

Notice that we only want to find the C−1 , C0 , and C1 terms in the Laurent Series. We may now
 just look at the terms in f (t) and see that each term (1) to (4) will be:
i 1 i
(1c) = − t − t2 + ...
22 2     
1 i 3 2i 1 11
(2c) = − + − + t− + t2 + ...
25 50 250 125 625 1250
 
1 i 1
(3c) = +
5 10 t
     
1 i 1 i 1 i
(4c) = − + + − t− + t2 + ...
20 20 20 40 80 40

Combining terms of powers −1,0,and 1 we get:

 
1 i
C−1 = +
5 10
 
9 19i
C0 = − −
100 50
 
231 41i
C1 = − +
500 1000

Below are some plots of the function and its Laurent series near z = i.

2. Describe all the singularities (none, removable, pole (if so, what order), essential, etc.) in the finite
plane (i.e. no need to consider z = ∞) of the following functions:

(a) f (z) = (cossinz)−1

z
We know that 1/sin(z) has poles at z = n, n = 0, ±1, ±2, .... We notice however
that cos(z) − 1 is zero for z = 2n, n = 0, ±1, ±2, .... We therefore conclude that we have poles
at z = (2n − 1) and removable singularities at z = 2n, n = 0, ±1, ±2, .... The poles are of order
1.
1/2
(b) f (z) = z z+1+1 We notice that f (z) could have a pole at z = −1. We want to make sure that there
really is a pole there. If we were to get a zero in the numerator as well as in the denominator,
we would actually have a removable singularity, not a pole. We must must be careful about the
multivalued nature of z 1/2 . So let us check:
±i + 1 0
f (−1) = 6=
0 0
We can therefore conclude that we have a pole of order 1 at z = −1. To compute the residue
1/2 1/2 1/2 +1 ±i(1− 12 (z+1)+...)+1
and strength, expand z z+1+1 around z = −1. z z+1+1 = ((z+1)−1)
z+1
= z+1
=
±i+1
z+1
+ a0 + a1 (z + 1) + ...
 Figure 3: Magnitude of f(z) and then of the Laurent Series

1/2
(c) f (z) = z z−1−1 We first notice that we could have a pole at z = 1. However, we must again check
the numerator for a potential 0 (which would result in a removable singularity).
±1 − 1 0 −2
6= or
f (1) =
0 0 0
This is very interesting as you get a pole if you pick the negative branch of the square root,
and you get a removable singularity if you pick the positive branch. The pole is of order 1.
1/2 1/2 1/2 −1
To compute the residue and strength, expand z z−1−1 around z = 1. z z−1−1 = ((z−1)+1)
z−1
=
±(1+ 12 (z−1)+...)−1 ±1−1
z−1
= z−1
+ a0 + a1 (z + 1) + ...

3. Let t be a parameter, and consider the analytic function

1
f (z, t) = et(z− z )/2

(a) Tell where in the complex z-plane f (z, t) has singularities. f (z, t) has singularities at z = 0 and
at z = ∞. The Laurent expansion of f (z, t) (in terms of z) around the origin will naturally
 have coefficients that depend on the parameter t:
∞
1
X
et(z− z )/2 = Jn (t)z n (1)
n=−∞

We want to check that the Jn (J-Bessel functions) satisfy the following ODE:
d2 y dy
t2 + t + (t2 − n2 )y = 0 (2)
dt2 dt
We can do so by checking that f(z, t) satisfies the following ODE (Derived from (2) and f(z, t)):
2
 
2∂ f ∂f 2 ∂ ∂f
t +t +t f =z z (3)
∂t2 ∂t ∂z ∂z
Because the Jn (t) are uniquely determined by f (z, t) (they are the Laurent Coefficients), If
f (z, t) does indeed satisfy (3), then the Jn (t) must satisfy (2). We shall check (3) by direct
differentiation, equating left and right:
2 t
t2 e 2 (z−1/z) t(2z(z 2 − 1) + t(1 + z 2 )2 )
  
1 t
(z−1/z) t 1 t
(z−1/z) 2 t
(z−1/z)
z− e2 + z− e2 + t e2 =
4 z 2 z 4z 2
... = ...
0 = 0
We have shown that (3) is satisfied by f (z, t) and thus, (2) is satisfied by Jn (t).
(b) Jn (t) being the Laurent coefficients of f (z, t), we know that, :
˛   ˛  t(z−1/z)/2 
1 f (z, t) 1 e
Jn (t) = dz = dz
2πi C (z − z0 )n+1 2πi C z n+1
˛ iθ −1 −iθ
!
1 et(re −r e )/2
= dz
2πi C (reiθ )n+1

At this point, our contour C is arbitrary as long as it encloses the origin. For convenience, we
pick C to be the unit circle.
ˆ π  it sin θ 
1 e iθ
Jn (t) = ie dθ
2πi −π (eiθ )n+1
ˆ π  it sin θ 
1 e
= dθ
2π −π (eiθ )n
ˆ π
1
e−i(nθ−t sin θ) dθ


=
2π −π
ˆ π
1
= (cos(nθ − t sin θ) − i sin(nθ − t sin θ)) dθ
2π −π
ˆ
1 π
= (cos(nθ − t sin θ)) dθ
π 0
 ´ ∞ dx
4. Use calculus of residues to evaluate −∞ 1+x 4 . Let us extend this real integral to the complex plane.
´ ∞ dz
The integral becomes: −∞ 1+z4 . We would like to use residue calculus, that is, we need some sort of
1
closed contour that encloses singularities of the function. We first note that 1+z 4 has 4 poles of order
√ √ √ √ √ √ √ √
1, located at z1 = ( 2 + i 2)/2, z2 = −( 2 + i 2)/2, z3 = −( 2 − i 2)/2 and z4 = ( 2 − i 2)/2.
Note that the current integration path is not a closed contour, it simply runs from z =?∞ to z = ∞
If we were to extend the present contour in the upper (or lower) half plane with an arc, we would
then have a closed contour. We need however to ensure that adding the upper or lower arc to our
path of integration does not change the integral value. To test if the path to add will not change the
integral value, we use the extremely useful estimation:
˛ 
 
 f (z)dz  ≤ M · L
 
C

where M is the upper bound of |f (z)| along C, and L is the Length of C. In our case, let us try to
estimate the magnitude of the integral along the proposed arc in the upper half plane. Such an arc
would have a radius R, with R → ∞. Then,
 
 1  1 1
M ∝ 
4
∝
4
∝ 4
1+z  R −1 R

while, L = πR. Yielding: ˛ 

 1  π
 1 + z 4 dz  ≤ R3 → 0
 
C

This estimation tells us that the extra path in the upper half plane actually did not change the value
of the original integral. The trick now is that we now have a closed path, enclosing two singularities,
z1 and z2 . We can now use Residue calculus. We know that:
˛ X
f (z)dz = 2πi Res
C
P
where Res is the sum of the Residues of the singularities enclosed by C. To calculate the Residue
of z1 and z2 , notice that those points are simple poles and that f (z) = N (z)
D(z)
, where N (z0 ) 6= 0, z0 the
location the pole. We can then use the formula:

N (z0 )
Resz=z0 =
D0 (z0 )

Applying this equation to both of our enclosed poles yields:

e−3iπ/4
Resz=z1 =
4
And
e−9iπ/4
Resz=z2 =
4
 Now we can add everything up:
˛ ˆ ∞
π
f (z)dz = f (z)dz = √
C −∞ 2

5. The following three figures (labeled a, b and c; with the real axis pointing towards the right and
slightly upwards) illustrate three of the six functions f (z) that are listed below. Match each of the
pictures with its corresponding function, and also tell whether the figure shows Re f (z), Im f (z), or
|f (z)| . Explain in each case some characteristic of the figures which allows you to eliminate all other
possibilities than the one you decided on.

• We can see that the function in Figure 4 has a branch cut on the negative real axis. This
feature eliminates all the choices except for f (z) = z 1/2 . Now we must determine which part
of the function is plotted. Let us write out f (z) = z 1/2 = (reiθ+2inπ )1/2 = r1/2 eiθ/2+inπ . We
can immediately rule out the Magnitude of f (z) because it would be independent of θ, which
is clearly not the case from the plot. So we try the real part: r1/2 cos(θ/2 + nπ). The extra nπ
is what is giving us multivaluedeness, and thus why there is a branch cut on the negative real
axis. We can omit it now for analysis of the function behavior now that the function is single
valued (thanks to the branch cut). We note that for θ = −π, r1/2 cos(θ/2 + nπ) = 0 and the
same thing for θ = π. This is what is seen on the plot. The imaginary part would be the sine
function and at those angle√ would be equal to ±1. We can then conclude that the plot in Figure
4 is the Real part of the z.
• We immediately notice that we are dealing with the absolute value of some function. We also
notice the periodicity in the pole locations. We therefore think of one of the trig functions. Let
1
us try coshz .

1 2
f (z) = = z
cosh z e + e−z
2
= x iy
e e + e−x e−iy
We can see that evaluating on the line x = 0 we then have:
2 2
−iy
=
eiy +e cos y

so we will have poles for y = (2n − 1)π/2, n an integer. This is indeed what we see on the
graph. We may now realize that f (z) is the π/2 rotated version of 1/cosz.
• We can tell this plot has no singularities, at least in the region shown. We can also notice that
the function has a saddle point at the origin. We are only left with z 2 and z 3 as choices. Let us
expand z 2 :
f (z) = z 2 = x2 − y 2 + 2ixy
 √
Figure 4: Re z

The imaginary part of this function indeed does have a saddle point at the origin. Furthermore,
it just looks like a quadratic on the line y = x and a minus quadratic on the line y = −x. This
property can also be shown to be the Imaginary part of z 2 as:

(2xy)y=x = 2x2

And
(2xy)y=−x = −2x2
We can therefore conclude that the plot in Figure 6 is the Imaginary part of z 2 .
  1 
Figure 5:  cosh z

Figure 6: Im (z 2 )