Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 1 pps

We substitute this into the second Cauchy-Riemann equation to determine ax... If f z is not analytic at z0, butlimz→z0f z exists, then the function has a removable singularity at z0... I

Result 8.3.1 If f (z) = u + ıv is an analytic function then u and v are harmonic functions That is, the Laplacians of u and v vanish ∆u = ∆v = 0 The Laplacian in Cartesian and polar coordinates is

r2

∂2

∂θ2 Given a harmonic function u in a simply connected domain, there exists a harmonic function

v, (unique up to an additive constant), such that f (z) = u + ıv is analytic in the domain One can construct v by solving the Cauchy-Riemann equations.

Example 8.3.1 Is x2 the real part of an analytic function?

The Laplacian of x2 is

∆[x2] = 2 + 0

x2 is not harmonic and thus is not the real part of an analytic function

Example 8.3.2 Show that u = e−x(x sin y − y cos y) is harmonic

∂u

∂x = e

−x

sin y − ex(x sin y − y cos y)

= e−xsin y − x e−xsin y + y e−xcos y

∂2u

∂x2 = − e−xsin y − e−xsin y + x e−xsin y − y e−xcos y

= −2 e−xsin y + x e−xsin y − y e−xcos y

∂y2 = e−x(−x sin y + sin y + y cos y + sin y)

= −x e−xsin y + 2 e−xsin y + y e−xcos yThus we see that ∂∂x2u2 + ∂∂y2u2 = 0 and u is harmonic

Example 8.3.3 Consider u = cos x cosh y This function is harmonic

uxx+ uyy = − cos x cosh y + cos x cosh y = 0Thus it is the real part of an analytic function, f (z) We find the harmonic conjugate, v, with the Cauchy-Riemannequations We integrate the first Cauchy-Riemann equation

v = − sin x sinh y + c

The analytic function is

f (z) = cos x cosh y − ı sin x sinh y + ıc

We recognize this as

f (z) = cos z + ıc

Example 8.3.4 Here we consider an example that demonstrates the need for a simply connected domain Consider

u = Log r in the multiply connected domain, r > 0 u is harmonic

We are able to solve for v, but it is multi-valued Any single-valued branch of θ that we choose will not be continuous

on the domain Thus there is no harmonic conjugate of u = Log r for the domain r > 0

If we had instead considered the simply-connected domain r > 0, | arg(z)| < π then the harmonic conjugate would

be v = Arg(z) + c The corresponding analytic function is f (z) = Log z + ıc

Example 8.3.5 Consider u = x3− 3xy2+ x This function is harmonic

uxx+ uyy = 6x − 6x = 0Thus it is the real part of an analytic function, f (z) We find the harmonic conjugate, v, with the Cauchy-Riemannequations We integrate the first Cauchy-Riemann equation

vy = ux = 3x2− 3y2+ 1

v = 3x2y − y3+ y + a(x)Here a(x) is a constant of integration We substitute this into the second Cauchy-Riemann equation to determine a(x)

vx = −uy

6xy + a0(x) = 6xy

a0(x) = 0a(x) = c

Here c is a real constant The harmonic conjugate is

Example 8.4.1 Consider f (z) = z3/2 The origin and infinity are branch points and are thus singularities of f (z) Wechoose the branch g(z) =√

z3 All the points on the negative real axis, including the origin, are singularities of g(z)

Removable Singularities

Example 8.4.2 Consider

f (z) = sin z

z .This function is undefined at z = 0 because f (0) is the indeterminate form 0/0 f (z) is analytic everywhere in thefinite complex plane except z = 0 Note that the limit as z → 0 of f (z) exists

= lim

z→0

1 − sin(z)/zz

= lim

z→0

sin(z)2

= 0

We call the point at z = 0 a removable singularity of sin(z)/z because we can remove the singularity by defining thevalue of the function to be its limiting value there

Consider a function f (z) that is analytic in a deleted neighborhood of z = z0 If f (z) is not analytic at z0, butlimz→z0f (z) exists, then the function has a removable singularity at z0 The function

Poles If a function f (z) behaves like c/ (z − z0)nnear z = z0 then the function has an nth order pole at that point.More mathematically we say

Another way to say that a function has an nth order pole is that f (z) is not analytic at z = z0, but (z − z0)nf (z)

is either analytic or has a removable singularity at that point

Example 8.4.3 1/ sin (z2) has a second order pole at z = 0 and first order poles at z = (nπ)1/2, n ∈ Z±

lim

z→0

z2sin (z2) = limz→0

2z2z cos (z2)

a removable singularity at infinity If limz→∞f (z)/zn= c 6= 0 then f (z) has an nth order pole at infinity.

Result 8.4.2 Categorization of Singularities Consider a function f (z) that has a larity at the point z = z0 Singularities come in four flavors:

singu-Branch Points singu-Branch points of multi-valued functions are singularities.

Removable Singularities If limz→z0 f (z) exists, then z0 is a removable singularity It

is thus named because the singularity could be removed and thus the function made analytic at z0 by redefining the value of f (z0).

Poles If limz→z0 (z − z0)nf (z) = const 6= 0 then f (z) has an nth order pole at z0.

Essential Singularities Instead of defining what an essential singularity is, we say what it

is not If z0 neither a branch point, a removable singularity nor a pole, it is an essential singularity.

A pole may be called a non-essential singularity This is because multiplying the function by an integral power of

z − z0 will make the function analytic Then an essential singularity is a point z0 such that there does not exist an nsuch that (z − z0)nf (z) is analytic there

8.4.2 Isolated and Non-Isolated Singularities

Result 8.4.3 Isolated and Non-Isolated Singularities Suppose f (z) has a singularity at

z0 If there exists a deleted neighborhood of z0 containing no singularities then the point is

an isolated singularity Otherwise it is a non-isolated singularity.

If you don’t like the abstract notion of a deleted neighborhood, you can work with a deleted circular neighborhood.However, this will require the introduction of more math symbols and a Greek letter z = z0 is an isolated singularity

if there exists a δ > 0 such that there are no singularities in 0 < |z − z0| < δ

Example 8.4.5 We classify the singularities of f (z) = z/ sin z

z has a simple zero at z = 0 sin z has simple zeros at z = nπ Thus f (z) has a removable singularity at z = 0and has first order poles at z = nπ for n ∈ Z± We can corroborate this by taking limits

lim

z→0f (z) = lim

z→0

zsin z = limz→0

1cos z = 1

lim

z→nπ(z − nπ)f (z) = lim

z→nπ

(z − nπ)zsin z

= lim

z→nπ

2z − nπcos z

R < |z| < ∞.) Thus z = ∞ is a non-isolated singularity

We could also determine this by setting ζ = 1/z and examining the point ζ = 0 f (1/ζ) has first order poles at

ζ = 1/(nπ) for n ∈ Z \ {0} These first order poles come arbitrarily close to the point ζ = 0 There is no deletedneighborhood of ζ = 0 which does not contain singularities Thus ζ = 0, and hence z = ∞ is a non-isolated singularity.The point at infinity is an essential singularity It is certainly not a branch point or a removable singularity It is not apole, because there is no n such that limz→∞z−nf (z) = const 6= 0 z−nf (z) has first order poles in any neighborhood

of infinity, so this limit does not exist

8.5 Application: Potential Flow

Example 8.5.1 We consider 2 dimensional uniform flow in a given direction The flow corresponds to the complexpotential

Φ(z) = v0e−ıθ0z,where v0 is the fluid speed and θ0 is the direction We find the velocity potential φ and stream function ψ

Φ(z) = φ + ıψ

φ = v0(cos(θ0)x + sin(θ0)y), ψ = v0(− sin(θ0)x + cos(θ0)y)These are plotted in Figure8.1 for θ0 = π/6

Figure 8.1: The velocity potential φ and stream function ψ for Φ(z) = v0e−ıθ0z

Next we find the stream lines, ψ = c

v0(− sin(θ0)x + cos(θ0)y) = c

v0cos(θ0) + tan(θ0)x

Figure 8.2: Streamlines for ψ = v0(− sin(θ0)x + cos(θ0)y).

Figure 8.2 shows how the streamlines go straight along the θ0 direction Next we find the velocity field

v = ∇φ

v = φxx + φˆ yyˆ

v = v0cos(θ0)ˆx + v0sin(θ0)ˆy

The velocity field is shown in Figure 8.3

Example 8.5.2 Steady, incompressible, inviscid, irrotational flow is governed by the Laplace equation We considerflow around an infinite cylinder of radius a Because the flow does not vary along the axis of the cylinder, this is atwo-dimensional problem The flow corresponds to the complex potential

Figure 8.3: Velocity field and velocity direction field for φ = v0(cos(θ0)x + sin(θ0)y).

We find the velocity potential φ and stream function ψ

These are plotted in Figure8.4

Figure 8.4: The velocity potential φ and stream function ψ for Φ(z) = v0 z + az2

Next we find the stream lines, ψ = c

r = c ±

p

c2+ 4v0sin2θ2v0sin θ

Figure 8.5 shows how the streamlines go around the cylinder Next we find the velocity field

Figure 8.5: Streamlines for ψ = v0 r − ar2 sin θ.

v = ∇φ

v = φrˆr + φθ

r

ˆθ

Figure 8.6: Velocity field and velocity direction field for φ = v0 r + ar2 cos θ.

8.6 Exercises

Complex Derivatives

Exercise 8.1

Consider two functions f (z) and g(z) analytic at z0 with f (z0) = g(z0) = 0 and g0(z0) 6= 0

1 Use the definition of the complex derivative to justify L’Hospital’s rule:

lim

z→z 0

f (z)g(z) =

Show that the following functions are nowhere analytic by checking where the derivative with respect to z exists.

1 sin x cosh y − ı cos x sinh y

1 Show that ez is not analytic.

2 f (z) is an analytic function of z Show that f (z) = f (z) is also an analytic function of z

2 Determine all points z where these functions are analytic

3 Determine which of the following functions v(x, y) are the imaginary part of an analytic function u(x, y)+ıv(x, y).For those that are, compute the real part u(x, y) and re-express the answer as an explicit function of z = x + ıy:

Hint, Solution

Exercise 8.13

Show that the logarithm log z is differentiable for z 6= 0 Find the derivative of the logarithm

Hint, Solution

w = u + ıv is an analytic function of z φ(x, y) is an arbitrary smooth function of x and y When expressed in terms

of u and v, φ(x, y) = Φ(u, v) Show that (w0 6= 0)

∂Φ

∂u − ı∂Φ

∂v =

 dwdz

dwdz

Hint, Solution

Exercise 8.16

Show that the functions defined by f (z) = log |z|+ı arg(z) and f (z) =p|z| eı arg(z)/2are analytic in the sector |z| > 0,

| arg(z)| < π What are the corresponding derivatives df /dz?

Hint, Solution

Exercise 8.17

Show that the following functions are harmonic For each one of them find its harmonic conjugate and form thecorresponding holomorphic function

1 u(x, y) = x Log(r) − y arctan(x, y) (r 6= 0)

2 u(x, y) = arg(z) (| arg(z)| < π, r 6= 0)

3 u(x, y) = rncos(nθ)

is differentiable and where it is analytic.

2 Evaluate the derivative of

f (z) = ex2−y2(cos(2xy) + ı sin(2xy))and describe the domain of analyticity

Hint, Solution

Exercise 8.19

Consider the function f (z) = u + ıv with real and imaginary parts expressed in terms of either x and y or r and θ

1 Show that the Cauchy-Riemann equations

ux = vy, uy = −vxare satisfied and these partial derivatives are continuous at a point z if and only if the polar form of the Cauchy-Riemann equations

ur = 1

rvθ,

1

ruθ = −vr

is satisfied and these partial derivatives are continuous there

2 Show that it is easy to verify that Log z is analytic for r > 0 and −π < θ < π using the polar form of theCauchy-Riemann equations and that the value of the derivative is easily obtained from a polar differentiationformula

3 Show that in polar coordinates, Laplace’s equation becomes

2 Construct functions that have the following zeros or singularities:

(a) a simple zero at z = ı and an isolated essential singularity at z = 1

(b) a removable singularity at z = 3, a pole of order 6 at z = −ı and an essential singularity at z∞

Hint, Solution

Hint 8.20

Hint 8.21

Hint 8.22

CONTINUE

g0(z0) = limz→z 0

f (z)g(z)This proves L’Hospital’s rule



z=ı

= 16

We calculate the complex derivative in the coordinate directions.

1 Consider f (x, y) = sin x cosh y − ı cos x sinh y The derivatives in the x and y directions are

These derivatives exist and are everywhere continuous We equate the expressions to get a set of two equations.

cos x cosh y = − cos x cosh y, sin x sinh y = − sin x sinh y

cos x cosh y = 0, sin x sinh y = 0



x = π

2 + nπ

and (x = mπ or y = 0)The function may be differentiable only at the points

x = π

2 + nπ, y = 0.

Thus the function is nowhere analytic

2 Consider f (x, y) = x2− y2 + x + ı(2xy − y) The derivatives in the x and y directions are

∂f

∂x = 2x + 1 + ı2y

−ı∂f

∂y = ı2y + 2x − 1These derivatives exist and are everywhere continuous We equate the expressions to get a set of two equations

We define g(z) = log(f (z)).

g (z1+ z2) = g (z1) + g (z2)This is a linear equation which has exactly the solutions:

g(z) = cz

Thus f (z) has the solutions:

f (z) = ecz,where c is any complex constant We can write this constant in terms of f0(0) We differentiate the original equationwith respect to z1 and then substitute z1 = 0

We integrate the first equation to obtain v = a + g(y) where a is a constant and g(y) is an arbitrary function Then

we substitute this into the second equation to determine g(y)

g0(y) = 0g(y) = b

We see that the imaginary part of f (z) is a constant and conclude that f (z) is constant

Constant Imaginary Part Next assume that f (z) has constant imaginary part We solve the Cauchy-Riemannequations to determine the real part

ux = vy, uy = −vx

ux = 0, uy = 0

We integrate the first equation to obtain u = a + g(y) where a is a constant and g(y) is an arbitrary function Then

we substitute this into the second equation to determine g(y)

g0(y) = 0g(y) = b

We see that the real part of f (z) is a constant and conclude that f (z) is constant

Constant Modulus Finally assume that f (z) has constant modulus



= 0

This system has non-trivial solutions for u and v only if the matrix is non-singular (The trivial solution u = v = 0 isthe constant function f (z) = 0.) We set the determinant of the matrix to zero.

uxvy − uyvx = 0

We use the Cauchy-Riemann equations to write this in terms of ux and uy

u2x+ u2y = 0

ux = uy = 0Since its partial derivatives vanish, u is a constant From the Cauchy-Riemann equations we see that the partialderivatives of v vanish as well, so it is constant We conclude that f (z) is a constant

Constant Modulus Here is another method for the constant modulus case We solve the Cauchy-Riemannequations in polar form to determine the argument of f (z) = R(x, y) eıΘ(x,y) Since the function has constant modulus

R, its partial derivatives vanish

Rx = RΘy, Ry = −RΘx

RΘy = 0, RΘx = 0The equations are satisfied for R = 0 For this case, f (z) = 0 We consider nonzero R

ux = vy, uy = −vx

for this problem.

f (z) is not analytic at the point z = 0 We show this by calculating the derivative

For most values of θ the limit does not exist Consider θ = π/4.

f0(0) = lim

∆r→0

er −4

∆r eıπ/4 = ∞Because the limit does not exist, the function is not differentiable at z = 0 Recall that satisfying the Cauchy-Riemannequations is a necessary, but not a sufficient condition for differentiability

Solution 8.8

1 We find the Cauchy-Riemann equations for

f (z) = R(r, θ) eıΘ(r,θ).From Exercise8.3 we know that the complex derivative in the polar coordinate directions is

We equate the derivatives in the x and y directions.

∂

∂xR eıΘ = −ı ∂

∂yR eıΘ(Rx+ ıRΘy) eıΘ = −ı (Rx+ ıRΘy) eıΘ

We divide by eıΘ and equate the real and imaginary components to obtain the Cauchy-Riemann equations

Rx = RΘy, Ry = −RΘx

Solution 8.9

1 A necessary condition for analyticity in an open set is that the Cauchy-Riemann equations are satisfied in thatset We write ez in Cartesian form

ez = ex−ıy = excos y − ı exsin y

Now we determine where u = excos y and v = − exsin y satisfy the Cauchy-Riemann equations

Thus we see that the Cauchy-Riemann equations are not satisfied anywhere ez is nowhere analytic

2 Since f (z) = u + ıv is analytic, u and v satisfy the Cauchy-Riemann equations and their first partial derivativesare continuous

f (z) = f (z) = u(x, −y) + ıv(x, −y) = u(x, −y) − ıv(x, −y)

is differentiable and where it is analytic.

2 Evaluate the derivative of

f (z) = ex2< /sup>−y2< /sup>(cos(2xy) + ı sin(2xy) )and describe the domain... nowhere analytic

2 Consider f (x, y) = x2< /small>− y2< /small> + x + ı(2xy − y) The derivatives in the x and y directions are

∂f

∂x = 2x + + ı2y

−ı∂f... data-page= "26 ">

Hint 8 .20

Hint 8 . 21

Hint 8 .22

CONTINUE

g0(z0)