# Chapter 2. Functions of a Complex Variable – Engineering Mathematics, Volume III

## Functions of a Complex Variable

#### 2.1 Introduction

Many engineering problems can be solved using Complex Variable Theory. Some of the elementary problems dealing with applications to electric circuits or mechanical vibrating systems need only elementary knowledge of complex analysis. Advanced problems in heat conduction, fluid flow and electrostatics require knowledge of complex analysis.

Equations such as x2 = –1 have no solutions in real variables. By defining i = as a complex unit, the system of real numbers can be extended to the complex number system.

#### 2.2 Complex Numbers-Complex Plane

A complex number z is an ordered pair of real numbers x and y written as z = (x, y). Here x is called the real part and y the imaginary part of z. We write

x = Re z,   y = Im z

Equality of two complex numbers

Two complex numbers z1 = (x1, y1) and z2 = (x2, y2) are equal if and only if their real parts are equal and their imaginary parts are equal, i.e.,

x1 = x2, y1 = y2

Imaginary unit

(0, 1) is called the imaginary unit and is denoted by i.

i = (0,1)          (2.1)

Addition of two complex numbers z1 = (x1, y1) and z2 = (x2, y2) is defined by

 z1 + z2 = (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2) (2.2)

Multiplication is defined by

 z1 · z2 = (x1,y1)(x2,y2) = (x1x2 − y1y2, x1y2 + x2y1) (2.3)

In particular, these two definitions imply that

(x1, 0) + (x2, 0) = (x1 + x2, 0) and
(x1, 0) (x2, 0) = (x1x2, 0)

as for real numbers x1 and x2. Hence the complex numbers extend the reals. We can write

 (x, 0) = x,   (0,y) = iy,   (x,y ∈ ℝ) (2.4) ∵ iy = (0, 1)(y, 0) = (0 · y − 1 · 0,0 · 0 + 1 · y) = (0, y).

Since (x, y) = (x, 0) + (0, y) = x + iy it has great practical advantage to write z = x + iy. If x = 0 then z = iy and is called a pure imaginary number. Also Eqs. (2.1) and (2.3) give

i2 = i · i = (0,1)(0,1) = (−1,0) = −1          (2.5)

By the standard notation, addition may be done as follows:

(x1 + iy1) + (x2 + iy2) = (x1 + x2) + i(y1 + y2)

Multiplication can be done in the usual way noting that i2 = –1

(x1 + iy1)(x2 + iy2) = (x1x2y1y2) + i(x1y2 + x2y1)

Real and imaginary parts

Let z1 = 3 – 4i;    z2 = –5 + i.

Then Re z1 = 3    Im z1 = –4

Re z2 = –5      Im z2 = 1

Sum and product

 z1 + z2 = (3 − 4i) + (−5 + i) = −2 − 3i z1z2 = (3 − 4i)(−5 + i) = (−15 + 4) + i(3 + 20) = −11 + 23i

Subtraction and division

Subtraction and division are inverse operations of addition and multiplication.

Let z1 = (x1, y1) and z2 = (x2, y2). The difference z1z2 is the complex number z = (x, y) such that

 z + z2 = z1 ⇒ (x,y) + (x2,y2) = (x1,y1) ⇒ (x + x2,y + y2) = (x1,y1) ⇒ x + x2 = x1, y + y2 = y1 ⇒ (x,y) = (x1 − x2,y1 −y2).

 (x + iy) + (x2 + iy2) = (x1 + iy1) ⇒ x + x2 = x1, y + y2 = y1 ⇒ x = x1 − x2, y = y1 − y2 ⇒ z = z1 − z2 = (x1 − x2) + i(y1 − y2).

The quotient z1 / z2 (z2 ≠ 0) is the complex number z such that zz2 = z1

 (x + iy)(x2 + iy2) = (x1 + iy1) ⇒ xx2 − yy2 = x1, xy2 + yx2 = y1 xx2 − yy2 = x1 ∣x2 ∣y2 xy2 + yx2 = y1 ∣y2 ∣ x2 = x1x2 + y1y2 = x2y1 − x1y2

Solving z = (x, y)

In practice, we do as follows:

Figure 2.1

Difference and quotient

Let z1 = 3 − 2i and z2 = 4 + i

z = z1z2 = (3 − 2i) − (4 + i) = −1 − 3i

Complex numbers satisfy

1. closure
2. commutative
3. associative
4. existence of identity and
5. existence of inverse

laws w.r.t the operations of addition and multiplication and also the distributive laws as real numbers and form a field.

ℚ, ℝ, ℂ are fields; ℚ,ℝ satisfy order relation ‘<’; ℝ is complete ordered field. No such order relation exists in ℂ.

There is one-one correspondence between the set of reals ℝ and points in a line, and the set of complex numbers ℂ and the points in a plane.

Geometrical Representation of Complex Numbers

Complex plane or Argand plane

Choose two perpendicular lines, the horizontal line called the real axis and the vertical line called the imaginary axis. Thus we have a cartesian coordinate system. Plot a given complex number z = (x,y) = x + iy as a point P with coordinates x and y.

Addition and subtraction can be visualized as shown above (Fig. 2.1)

Complex conjugate numbers

The complex conjugate of a complex number z = x + iy is defined by

= xiy

Replace i by –i in the complex number to get its conjugate. It is obtained geometrically by reflecting the point z in the real axis.

Figure 2.2

We have z = x2 + y2; z + = 2x; z = 2iy

If z is real, z = x, then = z. Conversely if = z then z is real and z = x.

1. Conjugate of sum = Sum of conjugates

2. Conjugate of difference = Difference of conjugates

3. Conjugate of product = Product of conjugates

4. Conjugate of quotient = Quotient of conjugates

Example 2.1

Let z1 = 2 − 3i, z2 = 4 + 5i

##### EXERCISE 2.1
1. Powers of the imaginary unit:

Show that i2 = −1, i3 = −i, i4 = 1, i5 = i, ⋯ and

2. Multiplication by i is geometrically a counterclockwise rotation through π/2 (90°). Verify this by plotting z and iz and the angle of rotation for z = 4 + 2i, z = –1 + i, z = 5 – 2i.

Figure 2.3

3. Let z1 = –2 + 3i, z2 = 1 + i. Find
1. z1z2 = (–2 + 3i)(1 + i)

= (−2 − 3) + i(3 − 2) = −5 + i

2. Re(z1)2 = Re(5 – 12i) = 5 (Re z1)2 = (–2)2 = 4 z = x + iy
3. (1 + i)2 = 2i; Re(1 + i)2 = 0; (1 + i)4 = (2i)2 = −4

Polar Form of Complex Numbers, Powers and Roots

If we put

x = r cos θ, y = r sin θ

z = x + iy takes the polar form

z = r(cos θ + i sin θ) or re.

r is called the absolute value or modulus of z and is denoted by ∣z∣. ∣z∣ = r =

Figure 2.4

Geometrically, ∣z∣ represents the distance of the point from the origin.

The inequality ∣z1∣ > ∣z2∣ means that the point z1 is farther away from the origin than the point z2, and ∣z1z2∣ is the distance between the points z1 and z2. θ is called the argument or amplitude of z denoted by arg z or amp z.

Geometrically, θ is the directed angle from the positive x-axis to OP. For z = 0, this angle θ is undefined. For z ≠ 0 it is determined only up to an integer multiple of 2π. The value of θ that lies in the interval –π < θπ is called the principal value of the argument of z ≠ 0 and is denoted by Arg z with capital A. Thus –π < Arg zπ.

Example 2.2

Arg z = (principal value).

Since for any complex number z = x + iy,

Figure 2.5

Similarly |z| ≥ |y|

∴ |Re z| ≤ |z|; |Im z| ≤ |z|.

Let z1 and z2 be complex numbers. Then the origin and the points z1 and z1 + z2 are the vertices of a triangle whose sides are |z1|, |z2| and |z1 + z2|. We obtain triangle inequality

|z1 + z2| ≤ |z1| + |z2|.

Also, we have

||z1| − |z2|| ≤ |z1z2|

(The difference of two sides of a triangle cannot exceed the third side – Fig. 2.6.)

Figure 2.6

Multiplication and Division in Polar Form

 Let z1 = r1 (cos θ1 + i sin θ1), z2 = r2 (cos θ2 + i sin θ2) z1z2 = r1r2 [(cos θ1 cos θ2 − sin θ1 sin θ2) +i(sin θ1 cos θ2 + cos θ1 sin θ2)] = r1r2cis (θ1 + θ2) ∣z1z2∣ = ∣z1 ∣∣z2∣

arg (z1z2) = arg z1 + arg z2, up to multiple of 2π

arg (zz2) = arg z + arg z2 = arg z1

arg = arg z1 − arg z2, up to multiple of 2π

Example 2.3

Division

The quotient z = z1 /z2 is the number z satisfying zz2 = z1.

Hence |zz2| = |z||z2| = |z1|. arg zz2 = arg z + arg z2 = arg z1. This gives |z1/z2|=|z1|/|z2| (z2 ≠ = 0) and arg(z1/z2) = arg z1 − arg z2.

Integer Powers: DeMoivre’s Formula

zn = rn(cos + i sin )    n = 0, 1, 2, …

DeMoivre’s formula

(cos θ + i sin θ)n = cos + i sin

Put n = 2

cos2 θ − sin2 θ + 2i sin θ cos θ = cos 2θ + i sin 2θ

⇒ cos 2θ = cos2 θ − sin2 θ and sin 2θ = 2 sin θcos θ

Roots

z = wn (n = 1,2,3,…) nth root of z

Remark

This symbol is multi-valued (n-valued) in contrast to the usual convention in real calculus.

z = r(cos θ + i sin θ) w = R(cos φ + i sin φ)
wn = Rn (cos sin ) = z = r (cos θ + i sin θ)

By equating absolute values root is real positive and unique.

By equating arguments = θ + 2

thus

For these values of k, we get n distinct values of w; these values repeat themselves for higher values of k.

Consequently, for z ≠ 0 has the n district values

where k = 0, 1, … (n – 1). These n values, lie on a circle of radius with centre at the origin and constitute the vertices of a regular polygon of n sides.

The value of obtained by taking the principal value of Arg z and k = 0 in Eq. (2.6) is called the principal value of w = In particular, taking z = 1 we have |z| = r = 1 and Arg z = 0. Then Eq. (2.6) gives

These n values are called the nth roots of unity. They lie on the circle of radius 1 and centre 0, briefly called the unit circle (see Fig. 2.7).

If w denotes the value corresponding to k = 1 in Eq. (2.7) the n values of can be written as

1, w, w2, ⋯ wn−1

Similarly if w1 is any nth root of an arbitrary complex number z then the n values of in Eq. (2.6) are

w1, w1 w, w1 w2, ⋯ w1 wn−1

because multiplying w1 by wk corresponds to increasing the argument of w1 by 2kπn.

Figure 2.7

Definitions of Certain Terms

Distance between two points

If ‘z’ and ‘a’ are points in the complex plane ℂ the distance between them is d(z, a) = |za|

Set of points

By a set of points we mean a collection of points in the complex plane ℂ.

Example 2.4

Circle : S1 = {z ∈ ℂ | |iza| = ρ} is called a circle with centre ‘a’ and radius. ρ.

Right half-plane

S2 = {z ∈ ℂ|Re z > 0} is the right half-plane.

Neighbourhood of a jroint z0

The set of all points z for which |zz0| < δ where δ is a positive constant, is called a neighbourhood of z0 and written nbd of z0.

Deleted nbd of z0

The set of all points z for which 0 < |zz0| < δ is called a deleted nbd of z0.

Open circular disc

The set of all z for which |zz0| < ρ is called an open circular disc.

Closed circular disc

The set of all z for which |zz0| ⩽ ρ is called a closed circular disc.

Open annulus or circular ring

The set of all points z such that ρ1 < |zz0| < ρ2 (ρ1 < ρ2) is called an open annulus.

Closed annulus or circular ring

The set of all points z such that ρ ⩽ |zz0| ⩽ ρ2 (ρ1 < ρ2) is called a closed annulus.

Interior point of a set S

A point z0 is called an interior point of a set S if there is a nbd of z0 containing only the points of S.

Open set

A set S is called open if every point of S has a nbd every point of which belongs to S.

Example 2.5

1. The points inside a circle.
2. The points inside a square.
3. Right half-plane Re z = x > 0.

Open half-plane

Open upper half-plane:   The set of points z such that Im z = y > 0.

Open lower half-plane:   The set of points z such that Im z = y < 0.

Closed set

A set S is said to be closed if its complement S’ is open.

Example 2.6

The set of points on and inside the unit circle |z| = 1.

Bounded set

A set S is called bounded if all its points lie within a circle of sufficiently large radius.

Example 2.7

Points in a rectangle, ellipse or polygon.

Note:

Points on a straight line or in an infinite strip do not form a bounded set.

Connected Set

A set S in which any pair of points can be joined by a polygonal arc entirely lying in S, is called a connected set.

Figure 2.8

Domain

An open connected set is called a domain.

Example 2.8

Interior of a circle or a square or a closed polygon.

Boundary point of a set S

A point every nbd of which contains both points in S and points not in S.

Example 2.9

Points on the two circles C1 and C2 of an annulus.

Note:

If S is open, no boundary point belongs to S. If S is closed, every boundary point belongs to S. |z| ⩽ 1: closed, bounded set.

Im z2 = xy > 0 : open, unbounded set.

ρ1 < |zz0 < ρ2: open, bounded, unconnected set.

Region

A set of points consisting of a domain and some or all of its boundary points.

#### 2.2.1 Complex Function

Let S be a set of complex numbers. A function f defined on S is a rule that assigns to every z in S a complex number w, called the value of f at z. We write

w = f (z)

z varies in S and is called a complex variable and S is called the domain of definition of f.

z = x + iy, w = u(x, y) + iv(x, y)

Example 2.10

w = f (z) = z2 + 2z

 Let z = x + iy; z2 = (x + iy)2 = x2 − y2 + 2ixy; 2z = 2x + i2y ∴ w = u (x, y) + iv (x, y) = x2 −y2+2x+i (2xy+2y) Re w = u(x,y) = x2 − y2 + 2x; Im w = 2xy + 2y

#### 2.2.2 Limit of a Function

Let f (z) be a function defined in some deleted nbd of z0 and l be a complex number. If for every ∈> 0 there corresponds a δ > 0 such that

0 < ∣zz0 ∣ < δ ⇒ f(z) − l∣ < ∈

then we say that f (z) tends to l as z tends to z0. We write f (z) = l.

#### 2.2.3 Continuity at z0

Let f (z) be defined in a nbd of z0. If f (z) = f (z0) then we say that f is continuous at z0.

Figure 2.9

#### 2.2.4 Differentiability

A complex function f (z) is said to be differentiable at a point ‘z0’, if the limit

exists. The limit f’(z0) is called the derivative off (z) at z0. Putting Δz = zz0 we can write the limit as follows:

Remark

In real calculus, there are two directions – right-side and left-side – to approach a point x0. The exists ⇔

1. right-side limit exists
2. left-side limit exists
3. both are equal.

But in the case of complex numbers the situation is different. One can approach a point z0 in the complex plane from any of the infinite number of directions. The above limit must exist and be the same when we approach z0 in any of these directions. Thus the condition of differentiation is stringent in complex variables. Rules of differentiation are the same as in real calculus. All the complex-valued functions whose real-valued counterparts are differentiable are differentiable.

It is important to note that many simple functions do not have derivatives at any point.

Example 2.11

is not differentiable at any point z.

Solution     Let f (z) = = xiy. Writing Δz = Δx + iΔy

If Δy = 0 this is + 1 and if Δx = 0 this is –1. Thus Eq. (1) approaches +1 along path I but –1 along path II. Hence the limit of Eq. (1) as Δz → 0 does not exist at any point z.

Figure 2.10

Example 2.12

Show that f (z) = Re z = x is not differentiable at any z.

Solution

Example 2.13

f (z) = |z|2 is not derivable at any z except at z = 0.

Solution

Case (i) Δy → 0, Δx → 0

Case (ii) Δx → 0, Δy → 0

f′(z) exists ⇔ z + = −z + ⇒ 2z = 0 ⇒ z = 0 f′(z) does not exist though f (z) = |z|2 is continuous at all z (for z ≠ 0).

#### 2.2.5 Analytic Functions: Definition of Analyticity

A function f (z) is said to be analytic in a domain D if f (z) is defined and differentiable at all points of D.

The function f (z) is said to be analytic at a point z = z0 in D if f (z) is analytic in a nbd of z0 in D.

By an analytic function we mean a function that is analytic in some domain.

‘Regular’ and ‘holomorphic’ are other terms used for analytic functions. The fact that a function f (z) is holomorphic in a domain D is expressed as

fH(D).

#### 2.2.6 1Cauchy–2Riemann Equations

The Cauchy–Riemann equations (C.R.Es) provide us a test for the analyticity of a complex function

w = f (z) = u(x, y) + iv(x, y)

f is analytic in a domain D ⇔ (i) the first partial derivatives of u, v exist and are continuous in D, and (ii) u, v satisfy Cauchy–Riemann equations ux = vy, uy = –vx in D.

Example 2.14

f (z) = z2 = x2y2 + i2xy is analytic for all z

u = x2y2, v = 2xy satisfy CREs ux = 2x,

uy = −2y, vx = 2y, Vy = 2xux = vy uy = −vu

#### 2.2.7 Cauchy–Riemann Equations in Cartesian Coordinates

Theorem 2.1     If f (z) = u(x, y) + iv(x, y) is differentiable at z then at this point the first order partial derivatives of u and v exist and satisfy the CREs

ux = vy; uy = −vx          (2.8)

Proof:    By hypothesis f is differentiable at z. So

Hence the limit value along Paths I and II must be the same.

Figure 2.11

z = x + iy,    Δz = Δx + iΔy

Choose Path I. We let Δy → 0 first and then Δx → 0. After Δy = 0, Δz = Δx, then Eq. (2.10) becomes

f′(z) = ux + ivx          (2.11)

Similarly along Path II, we let Δx → 0 first and then Δy → 0. After Δx = 0, Δz = iΔy, then Eq. (2.10) becomes

∴ Existence of f(z) implies the existence of ux, vx, uy, vy in Eqs. (2.11) and (2.12).

Equating the real and imaginary parts of Eqs. (2.11) and (2.12) we get the CREs.

Theorem 2.2    If two real-valued functions u(x,y), v(x,y) of two real variables x and y have continuous first partial derivatives that satisfy CREs in some domain D then the complex function f (z) = u(x,y) + iv(x,y) is analytic in D.

Proof:    It is given that are continuous.

We have,

where ∈1→ 0, η1 → 0 as Δx → 0 and Δy → 0.

Again it is given that are continuous.

So we have,

where ∈2→ 0, η2 → 0 as Δx → 0 and Δy → 0.

∴ Δw = Δu + iΔv

where ∈ = ∈1 + i2 → 0, η = η1 + iη2 → 0 as Δx → 0 and Δy → 0.

Using CREs, we can write Eq. (2.15) as

Dividing both sides by Δz = Δx + iΔy and taking limit as Δz → 0

So, the derivative exists and is unique.

f (z) ∈ H(D).

Alternative forms of CREs

Theorem 2.3     (Complex form of CREs). Let f (z) = u(x, y) + iv(x, y) be differentiable in a domain D. Then the CREs ux = vy and uy = –vx can be put in a complex form as

fx = –ify.

Proof:    Let

 f (z) = u(x,y) + iv(x,y). Then, fx = ux + ivx,f = uy + ivy Now, fx = − ify ⇔ ux + ivx = −i(uy + ivy) ⇔ ux = vy and u, = −vx

#### 2.2.8 Cauchy-Riemann Equations in Polar Coordinates

Theorem 2.4     (CREs in polar coordinates). If f (z) = u(r, π) + iv(r, π) is differentiable at z = re then .

Also, .

Proof:     We know that x = r cos θ, y = r sin θ

and

Thus .

Similarly, we can prove that by

CREs.

Now,

CRES in Polar Coordinates [Alternative Method]

In polar coordinates, CREs are

Let x = r cos θ, y = r sin θ. Then z = x + iy = r cos θ + ir sin θ = re.

So, u + iv = f (z) = f (re)

Differentiating partially w.r.t r and θ, we get

Equating the real and imaginary parts

Note:

1. If f (z) = u(x, y) + iv(x,y) is analytic in a domain D then Re f = u = u(x,y) and Im f = v = v(x,y) have partial derivatives ux, uy, vx and vy and satisfy the partial differential equations ux = vy and uy = –vx called the Cauchy–Riemann Equations.
2. The derivative f ′(z) can be calculated using either of the two formulas:

f′(z) = ux + ivx or f′(z) = −iuy + vy.

3. Let f (z) = u + iv. Then the Cauchy–Riemann Equations ux = vy and uy = –vx are only necessary conditions for analyticity (differentiability) of f but not sufficient.
4. Let f (z) = u + iv. Then a sufficient condition for analyticity of f is that u and v have continuous partial derivatives and also satisfy CREs ux = vy and uy = –vx.

Properties of analytic functions

1. The sum, the difference, the product and the quotient of analytic functions are analytic. f, gH(D) ⇒f ±g,fg,f/g(g(z) ≠ 0) ∈ H(D).
2. An analytic function of an analytic function is analytic.
3. An entire function of an entire function is entire. (A function which is analytic in the entire complex plane is called an entire function, e.g. ez, polynomial, sin z, cos z etc.)
4. f ∈ H(D) ⇒ f is differentiable at all z ∈ D. ⇒ f is continuous at all zD.
5. If f is analytic at a point z = a then f is differentiable at z = a. However, the converse is not true.

Example 2.15

We have proved that f (z) = |z|2 is differentiable only at z = 0 and nowhere else. This function is not analytic at z = 0 since according to the definition of analyticity at a point, f must be differentiable not only at z = 0 but at every point of some nbd of 0.

Example 2.16

If f (z) = z2 is analytic for all z, then the CREs are satisfied.

For f (z) = = xiy we have u = x, v = −y

CRE(2) uy = 0 = − vx is satisfied

CRE(i) ux = 1 ≠ vy = − 1 is not satisfied.

is not analytic.

An analytic function of constant absolute value is constant

Example 2.17

Show that if f (z) is analytic in D and |f (z)| = k, a constant in D then f (z) is constant in D.

Solution    By assumption u2 + v2 = k2.

By differentiating partially w.r.t x and y we get

uux + vvx = 0, uuy + vvy = 0

⇒ (a) uuxvuy = 0 (b) uuy + vux = 0       (1)

∵ vx = −uy, vy = ux by CREs.

To eliminate uy, multiply 1(a) by u and 1(b) by v and add. We get

(u2 + v2)ux = 0.

Similarly to eliminate ux, multiply 1(a) by –v and 1(b) by u and add. We get

(u2 + v2)uy = 0.

If k2 = u2 + v2 = 0, then u = v = 0 ⇒ f = 0.

If k ≠ 0 then ux = uy = 0 ⇒ vx = vy = 0, by CREs.

u = constant, v = constant ⇒ f (z) is constant of D.

#### 2.2.9 Milne-Thomson’s Method

Example 2.18

Find the analytic function f (z) = u + iv where u(x,y) = ex (x cos yy sin y + cos y) + 2 cos x sinh y + 3x2 – 3y2.

Solution    Differentiating partially w.r.t x and y, we get,

ux = ex(x cos yy sin y + cos y) + 2 cos x sinh y + 3x2 − 3y2

uy = ex (− x sin y − sin yy cos y) + 2 sin x cosh y − 6xy + 1

We know that f ′(z) = ux + ivx = uxivy.

Replace x by z and y by 0

f′ (z) = ez(z · 1 − 0) + ez (1) + 0 + 3z2−0−2i sin zi

Integrating w.r.t z, we get

f (z) = 2ezez + ez + z3 + 2i cos ziz + c.

Limits, continuity, differentiability

Example 2.19

Find ‘a’ so that f is continuous.

Solution    f (z) is continuous everywhere except at z = – i.

Path I :

Path II :

by L’ Hospital’s rule

Thus = i = a = f (−i). Hence f is continuous if a = i, but not otherwise.

Example 2.20

Show that is not continuous at z=0.

Solution    Let z = x + iy.

Path I :

Along the x-axis:

Path II :

Along the y-axis:

does not exist. Hence f is not continuous at z = 0.

Example 2.21

Show that f (z) = continuous at all z.

Solution    We have |f (z) − f (z0)| = |0|. For any given ∈ choose ∈ = δ. We have |zz0| < δ ⇒ |f (z) − f (z0)| <∈⇒ f (z) is continuous at z0.

Since z0 is arbitrary f (z) is continuous for all z.

Example 2.22

Show that f (z) = |z| is continuous at all z.

Solution    We have |f (z) − f (z0)| = ⩽ |z − z0|. For any ∈> 0 choose ∈ = δ then |zz0| < δ ⇒ |f (z) − f (z0)| <∈⇒ f is continuous at z0.

Since z0 is arbitrary f is continuous at all z.

Example 2.23

Show that does not exist though the function approaches the same limit along every straight line through the origin.

Solution

Path I :

Path II :

Path III :

Along any straight line y = mx through (0, 0),

Path IV :

Along the parabolic path y = mx2

Along Path IV the limit is dependent on m.

Example 2.24

Solution

Path I :

Path III :

Since the limit does not exist f is discontinuous at z = 0.

Example 2.25

Show that every differentiable function is continuous.

Solution    Let f (z) be differentiable at ‘a’. Then

f (a) is well-defined. Consider

Hence f is continuous at z = a.

Remark

The converse of the above result is not true. That is, a function may be continuous but not differentiable.

Counter-examples:

The following functions are continuous at points specified against them but they are not differentiable at those points.

1. f (z) = at any point
2. g(z) = |z|2 at zero
3. h(z) = Re z at any point.

Example 2.26

Find the regions of analyticity of the functions

1. f(z) = ez
2. f(z) = (xy)2 + 2i(x + y)
3. f(z) = .

Solution

1. f(z) = ez = ex+iy = ex(cos y+sin y) = u+iv (say)

 ⇒ u = ex cos y, v = ex sin y ⇒ ux = ex cos y, uy = −ex sin y vx = ex sin y, vy = ex cos y ⇒ ux = vy; uy = −vx

ux, uy, vx, vy are continuous. CREs are satisfied for all x and y.

Hence ez is analytic throughout the complex plane and it is an entire function.

2.

 u = (x − y)2; v = 2(x + y) ux = 2(x − y); uy = −2(x − y); vx = 2, vy = 2 ux = vy ⇒ x − y = 1

also uy = −vxxy = 1

CREs are satisfied only along the straight line xy = 1.

f ′(z) = ux + ivx = 2(xy) + 2i = 2 + 2i exists along xy = 1 and not through any region (nbd) R. Hence f(z) is nowhere analytic.

3. f(z) = is the quotient of polynomials z – 2i and z2 + 1 which are analytic throughout ℂ. So, f(z) is analytic everywhere except at points where z2 + 1 = 0, i.e., z = ±i.

Example 2.27

Show that z2 is analytic for all z.

Solution

f(z) = u + iv = z2 = (x + iy)2 = (x2y2) + i (2xy)

u = x2y2,   v = 2xy

ux = 2x;   vy = 2x;   uy = –2y;   vx = 2y

ux = Vy,   uy = –Vx

ux, uy, vx, vy are continuous and CREs are satisfied; f(z) is analytic in ℂ.

Example 2.28

Show that f(z) = z + 2 is not analytic anywhere in the complex plane.

[JNTU 2000, 2003(2)]

Solution    Let z = x + iy and

f(z) = u + iv = z + 2 = (x + iy) + 2(xiy)

u = 3x, v = –y

ux = 3, Vx = 0, uy = 0, Vy = 1

CREs ux ≠ vy, uy = –vu are not satisfied.

Hence f(z) is not analytic anywhere.

Example 2.29

Show that = 0 where f(z) is an analytic function.

[JNTU 2002]

Solution    Since = – , we have

Since f(z) is analytic f(z) is analytic. f() is also analytic

Example 2.30

If f(z) = u + iv is an analytic function of z prove that

[JNTU 2003S (4)]

Solution    We know that

and

Differentiate first w.r.t taking f(z) as constant and then w.r.t z taking f() as constant; we have

Example 2.31

Find all values of k such that f(z) = ex(cos ky + i sin ky) is analytic.

[JNTU 2003(4)]

Solution

Let f (z) = u + iv = ex (cos ky + i sin ky)

 ⇒ u = ex cos ky, v = ex sin ky ⇒ ux = ex cos ky, uy = −kex sin ky vx = ex sin ky, vy = kex cos ky CREs. ux = vy ⇒ k = 1, uy = −vx ⇒ k = 1

f (z) is analytic ⇔ k = 1

Example 2.32

Prove that

where w = f(z) is analytic.

[JNTU 2004(4), 2004S, 2005]

Solution

Let f(z) = u + ivRe f(z) = u.

Also f = ux + ivx so that

Squaring

Now

Similarly

Now Eqs. (2) + (3) gives

We know that Re f(z) = u satisfies the Laplace’s equation so that

From Eqs. (4) and (5)

Example 2.33

Determine whether the function 2xy + i(x2y2) is analytic.

[JNTU 2003(4)]

Solution

Let f (z) = u + iv = 2xy + i(x2y2)

u = 2xy;     v = x2y2

CREs are ux = vy and uy = – vx which are clearly not satisfied. Hence the given function is not analytic.

Example 2.34

Show that f(z) = xy2(x + iy)/(x2 + y4) (z ≠ 0) and f(0) = 0 is not analytic at z = 0 although CREs are satisfied at the origin.

Solution

Clearly,

Along the path y = mx,

Also, along the path x = my2,

Since the limit value depends on m and is different for different paths, the limit does not exist. Hence f(z) is not differentiable at z = 0. So, f(z) is not analytic at z = 0.

We now prove that CREs are not satisfied at z = 0.

Let

Also u(0, 0) = 0 and v(0, 0) = 0

∵ f (z) = 0 at z = 0

The CREs are satisfied at the origin z = 0. Consequently, f(z) is not analytic at (0, 0) though CREs are satisfied at z = 0.

Example 2.35

Show that the function f(z) = is not analytic at the origin, although Cauchy-Riemann equations are satisfied at the origin.

Solution    Let f(z) = u(x, y) + iv(x, y) = so that u(x,y) = and v(x,y) = 0. We compute the partial derivatives at the origin.

Hence, Cauchy–Riemann equations , are satisfied at the origin.

We now find the derivative at the origin.

The limit must exist along any path leading to the origin. If we allow z = (x, y) → (0, 0) along the straight line y = mx, we obtain

which depends on m and hence not unique. Hence f(0) does not exist.

Note:

In this problem, all exist at (x, y) = (0, 0) and also satisfy Cauchy–Riemann equations at the origin. Yet f′(0) does not exist because the partial derivatives are not continuous at the origin.

Example 2.36

Prove that the function f (z) defined by

is continuous and the Cauchy-Riemann equations are satisfied at the origin but f(0) does not exist.

[VTU 2001, JNTU 2000, Punjab 1999]

Solution

Also, f(0) = 0 by hypothesis so that f(z) = f(0), when x → 0, y → 0 and y → 0, x → 0.

Now, let x and y both tend to 0 simultaneously along the path y = mx

Thus, irrespective the manner in which z → 0 we have f(z) = 0 = f(0), which implies that the function is continuous at z = 0.

Also,

 u(0, 0) + f (0) = 0. v(0, 0)

Now, we complete partial derivatives for u, v at the

Clearly, Cauchy–Riemann equations , are satisfied at the origin.

Also,

f′(0) depends on m and is not unique. Therefore, f′(z) does not exist at the origin.

f(z) is not analytic at the origin even though it is continuous and satisfies Cauchy–Riemann equations at z = 0 because the partial derivatives are not continuous at the origin.

#### 2.2.10 Orthogonal Trajectories

The curves u(x, y) = c1 and v(x, y) = c2 where f = u + ivH(D) represents two families of curves in the xy-plane which are mutually orthogonal.

Suppose a member of

u(x, y) = c1          (2.16)

cuts a member of

v(x, y) = c2          (2.17)

at a point P(x0, y0).

By implicit differentiation of u(x, y) = c1 we get = slope of curves Eq. (1)

at P =

Similarly, differentiation of v(x, y) = c2 gives m2 = slope of curves (2) at P =

The product of the slopes of the curves at the intersection point P m1. m2 =

... ux = vy, uy = –vx by CREs.

Hence the family of curves u(x, y) = c1 and the family v(x, y) = c2 cut each other orthogonally.

Example 2.37

Find the orthogonal trajectories of

u(x, y) = x3yxy3 = c          (1)

Solution    The curves given by v(x,y) = constant will be the required orthogonal trajectories if f(z) = u + iv is analytic.

Differentiating (1) partially w.r.t. x and y

ux = 3x2yy3;     uy = x3 − 3xy2;

vy = ux = 3x2yy3, by Cauchy Riemann Eqs.

Integrating w.r.t y we get v =

Differentiating w.r.t x, we get

where c1 is an arbitrary constant. The required orthogonal trajectory is

Example 2.38

Prove that u(x, y) = x3 – 3xy2 = c1 and v(x, y) = 3x2yy3 = c2 cut each other orthogonally.

[JNTU 2003(3)]

Solution    Let a member of the family

u(x, y) = x3 − 3xy2 = c1         (1)

intersect a member of the family

v(x, y) = 3x2yy3 = c2 at P = (x0, y0)         (2)

Differentiating Eqs.(1) and (2), we get

Clearly the product of their slopes at P = –1 and hence the problem.

Example 2.39

Find the orthogonal trajectories of the family of the curves r2 cos 2θ = c.

Solution    Let u(r, θ) = r2 cos 2θ = c

on integrating v(r, θ) = r2 sin 2θ + c1

Differentiating v w.r.t r, we get

Also = constant.

∴ The orthogonal trajectories of curves u(r, θ) = r2 cos 2θ = c are the curves v(r, θ) = r2 sin 2θ = c.

##### EXERCISE 2.2
1. Verify CREs for the functions f(z)=
1. z2,
2. iz + ,
3. sin z.
2. Prove that the following functions are nowhere differentiable:
1. f(z) = |z|,
2. f(z) = Re ,
3. f(z) = 2x + ixy2.
3. Prove that the following functions are differentiable at every point:
1. f(z) = x2y2 – 2xy + i(x2y2 + 2xy)
2. f(z) = 2x – 3y + i(3x + 2y)
3. f(z) = iz + 2.
4. Prove that f(z) = cosx(cosh y + a sinh y)+ i sin x (cosh y + b sinh y) is analytic if a = b = –1.
5. Determine constants a, b and c such that the following functions are differentiable at every point:
1. f(z) = x + ay + i(bx + cy)
2. f(z) = ex cos ay + i. ex sin(y + b) + c.
6. Prove that the functions f(z) and f() are simultaneously analytic.

#### 2.3 Laplace’s Equation: Harmonic and Conjugate Harmonic Functions

The real part u(x, y) and the imaginary part v(x, y) of an analytic function f(z) = u(x, y) + iv(x, y) satisfy the most important differential equation in physics:

Laplace’s equation

(∇ is called nabla or del) (viz.) ∇2u = uxx + uyy = 0 in D2v = vxx + vyy = 0 in D.

Equation ∇2F = 0 occurs in gravitation, electrostatics, fluid flow, heat conduction and so on.

Theorem 2.5    (Laplace’s equation) If f(z) = u(x, y) + iv(x, y) is analytic in a domain D then u and v satisfy Laplace’s equation

2u = uxx + uyy = 0,          (2.18)

2u = vxx + vyy = 0          (2.19)

in D and have continuous second partial derivatives in D.

Proof:    Differentiating ux = vy w.r.t x and uy = – vx w.r.t y, we have

uxx = vyx   and   uyy = −vxy          (2.20)

The derivative of an analytic function is itself analytic ⇒ u, v have continuous partial derivatives of all orders.

In particular, vyx = vxy.

Adding Eq. (2.20) we obtain Eq. (2.18). Equation (2.19) is similarly proved.

Solutions of Laplace’s equation having continuous second-order partial derivatives are called harmonic functions and their theory is called Potential Theory.

Hence the real and imaginary parts of an analytic function are harmonic functions.

If two harmonic functions u and v satisfy CREs in a domain D they are the real and imaginary parts of an analytic function f in D. v is said to be a conjugate harmonic function of u in D and vice versa.

Let f(z) = u + iv be an analytic function whose real part u is known. We can find the imaginary part v and also the function f(z) by following either of the two methods:

Method 1: Exact Differential Method using CREs u is a harmonic function

Let

where by Eq. (2.22)

Now

 = −∇2u = 0 by Eq. (2.21)

is an exact differential equation.

Hence integrating Eq. (2.23) we can find v. Having known u and v, f(z) = u + iv can be determined.

Method 2: Milne–Thomson’s method

Since z = x + iy, = xiy we have

Considering Eq. (2.24) as a formal identity in z and and putting = z, we get

f (z) = u(z, 0) + iv(z, 0)          (2.25)

which is obtained from f (z) = u(x, y) + iv(x, y) by replacing x by z and y by 0.

Now

If we set and then,

f′ (z) = Φ1(x, y) − iΦ2(x, y)          (2.26)

Replacing x by z and y by 0 in Eq. (2.26), we obtain

f ′(z) = Ф1 (z, 0) – iФ2(z, 0)

⇒f (z) = [Ф1 (z, 0) – iФ2(z, 0)] + c1

where c1 is a complex constant.

Similarly if v(x, y) is given, we can find u such that u + iv is analytic.

where

replacing x by z and y by 0.

Integrating we get

where c2 is a complex constant.

Example 2.40

Verify that u = x2y2y is harmonic in C and find a conjugate harmonic function v of u.

Solution    ∇2u = 0 by direct calculation. Now ux = 2x, uy = –2y – 1. Hence, because of CREs a conjugate v of u must satisfy the following:

vy = ux = 2x;         (1)

vx = −uy = 2y + 1         (2)

Integrating Eq. (1) w.r.t y and differentiating the result w.r.t x, we get

Comparison with Eq. (2) shows that dh / dx = 1 ⇒ h(x) = x + c. Hence v = 2xy + x + c (c any real constant) is the most general conjugate harmonic of the given function u. The corresponding analytic function is

 f (z) = u + iv = x2 − y2 − y + i(2xy + x + c) = z2 + iz + ic

The conjugate of a harmonic function is uniquely determined up to an arbitrary real additive constant.

Harmonic Functions

Example 2.41

If u(x, y) and v(x, y) are harmonic functions in a region R, prove that the function is an analytic function.

[JNTU 2003S(1)]

Solution    u and v are harmonic functions

Let

Differentiating Eqs. (3) and (4) partially w.r.t x and again w.r.t y, we get

Now by Eq. (2)

by Eq. (1)

Thus U, V satisfy CREs and their partial derivatives Ux, Uy, Vx, Vy are continuous.

Hence is an analytic function.

#### 2.3.1 Harmonic and Conjugate Harmonic Functions

Example 2.42

Show that u = 2 log(x2 + y2) is harmonic and find its harmonic conjugate.

[JNTU 2006(1)]

Solution    u = 2 log(x2 + y2)

u is an harmonic function.

If v(x, y) is the harmonic conjugate of u(x, y) then

(using CREs)

Integrating we get

Example 2.43

Show that if u and v are conjugate harmonic functions then the product uv is a harmonic function.

Solution    u, v are conjugate harmonic functions

uxx + uyy = 0,   vxx + vyy = 0         (1), (2)

ux = vy,   uy = − vx             (3), (4)

 If Φ = uv then Φx = uvx + vux, Φxx = uvxx + 2uxvx + vuxx and Φyy = uvyy + 2uyvy + vuyy = uvyy − 2vxux + vuyy, by CREs

Now Фxx + Фyy = u(vxx + vyy) + v(uxx + uyy) = 0.

Hence Ф = uv is a harmonic function.

Example 2.44

Show that v(x, y) = – sin x sinh y is harmonic. Find the harmonic conjugate of v.

Solution    Differentiating v partially w.r.t x and y, we get

 vx = −cos x sinh y, vxx = sin x sinh y vy = −sin x cosh y, vyy = −sin x sinh y ∇2v = vxx + vyy = 0 ∴ v is harmonic.

To find u (harmonic conjugate of v), by CREs,

ux = vy = −sin x cosh y,         (1)

uy = −vx = cos x sinh y         (2)

Integrating Eq. (1) partially w.r.t x, we get

u(x, y) = cos x cosh y + h(y)         (3)

where h(y) is an arbitrary function of y

Differentiating Eq. (3) partially w.r.t y we get

uy = cos x sinh y + h′(y) = −vx = cos x sinh y

⇒ h′(y) = 0 ⇒ h(y) = constant

∴ u(x, y) = cos x cosh y + c

The required analytic function is

 f (z) = cos x cosh y + c + i(− sin x sinh y). = cos(x + iy) + c = cos z + c.

Example 2.45

Prove that u = 2xx3 + 3xy2 is harmonic and find its harmonic conjugate. Find also the corresponding analytic function.

Solution    Let u = 2xx3 + 3xy2

ux = 2 − 3x2 + 3y2, uxx = −6x,

uy = 6xy, uyy = 6x

u is harmonic

Let v be a harmonic conjugate of u so that f (z) = u + iv is analytic.

By CREs we have vy = ux = 2 – 3x2 + +3y2

Integrating w.r.t y we get

v = 2y − 3x2y + y3 + A(x)                     (1)

where A(x) is an arbitrary function of x.

Differentiating Eq. (1) partially w.r.t x, we get vx = –6xy + A′(x) = –uy = –6xyA′(x) = 0 so that A(x) = c a pure complex constant

⇒ v = 2y − 3x2y + y3 + c

 Now f(z) = (2x − x3 + 3xy2) + i(2y − 3x2y + y3) + ic = 2(x + iy) − [(x3 − 3xy2) + i(3x2y − y3)] + ic = 2z − z3 + ic.

Example 2.46

Show that v(x, y) = – sin x sinh y is harmonic. Find the harmonic conjugate of v.

Solution    Differentiating v(x,y) partially w.r.t x and y, we get

vx = − cos x sinh y,    vxx = sin x sinh y    (1)

vy = − sin x cosh y,    vyy = − sin x sinh y    (2)

v is a harmonic function in D

To find u, the harmonic conjugate of v:

By CREs ux = vy = − sin x cosh y,            3(a)

uy = − vx = cos x sinh y            3(b)

Integrating 3(a) partially w.r.t x, we get

u = cos x cosh y + h(y)                     (4)

where h(y) is an arbitrary function of y.

Now differentiating Eq. (4) w.r.t y we get

 uy = cos x sinh y + h′(y) = −vx = cos x sinh y     by Eq. 3(b) ⇒ h′(y) = 0 ⇒ h(y) = c (const.)

The required analytic function is

 f (z) = u(x,y) + iv(x,y) = cos x cosh y − i sin x sinh y + c = cosz + c

Milne-Thomson’s method

Example 2.47

Find the analytic function f (z) = u + iv where u(x, y) = ex(x cos yy sin y) + 2 sin x sinh y + x3 – 3xy2 + y.

Solution    Differentiating u(x, y) partially w.r.t x and y, we get

 ux = ex (x cos y − y siny + cosy) + 2 cos x sinh y + 3x2 − 3y2 uy = ex (−x sin y − siny − y cosy) + 2 sinx coshy − 6xy + 1

We know that f′(z) = ux + ivx = uxiuy,

By CREs vx = – uy

Replace x by z and y by 0 to obtain f′(z)

 ∴ f ′(z) = ez (z · 1 − 0 + 1) + 0 + 3z2 − 0 − i[ez (0 − 0 − 0) + 2 sinz · 1 − 6z 0 + 1] = ez (z + 1) + 3z2 − 2i sin z − i

Integrating, we obtain

f (z) = zez + z3 + 2i cos ziz + c.

Example 2.48

Show that u(x, y) = sin x cosh y + 2 cos x sinh y + x2y2 + 4xy is harmonic. Find the analytic function f (z) with u as its real part.

Solution    Differentiating u partially w.r.t x and y twice, we get

 ux = cos x cosh y − 2 sin x sinh y + 2x + 4y uxx = − sin x cosh y − 2 cos x sinh y + 2 uy = sin x sinh y + 2 cos x cosh y − 2y + 4x uyy = sin x cosh y + 2 cos x sinh y − 2 ⇒ ∇2u =

Hence u is harmonic.

 Let Φ1(x,y) = ux and Φ2 (x,y) = uy so that Φ1 (z, 0) = cosz cosh 0 − 2 sinz sinh 0 + 2z = cos z + 2z

 and Φ2 (z, 0) = sin z sinh 0 + 2 cos z cosh 0 − 2 × 0 + 4z = 2 cos z + 4z

by Milne–Thomson’s method, where c is an arbitrary complex constant

Example 2.49

If u(x,y) = sin 2x/(cosh 2y + cos 2x) is the real part of analytic function f (z) = u(x, y) + iv(x, y). Find f (z).

Solution    We can easily verify that uxx + uyy = 0 and hence u is harmonic. Differentiating u partially w.r.t x and y we get

Example 2.50

Find the analytic function f (z) = u + iv if

Solution

Since f (z) = u + iv is an analytic function

ux = vy, uy = –vx(CRES)

Replacing x by z and y by 0,

Replacing x by z and y by 0,

u, (z, 0) + ux (z, 0) = 0                     (5)

Adding Eqs. (4) and (5), we get

Subtracting Eqs. (4) from (5), we get

 Now f (z) = u(z, 0) + iv(z, 0) ⇒ f ′(z) = ux (z, 0) + ivx (z, 0) ⇒ f ′(z) = ux (z, 0) − ivy (z, 0) = (1 + i)cosec2 z by Eqs. (6) and (7).

Integrating w.r.t z, we obtain

f(z) = (1 + i) cot z + c

where c is an arbitrary complex constant.

Example 2.51

If u + v = (xy)(x2 + 4xy + y2) find the analytic function f (z) = u + iv.

Solution

 f(z) = u + iv ⇒ (1 + i)f (z) = (u − v) + i(u + v) = U + iV (say) Now       V = u + v = (x − y)(x2 + 4xy + y2)

By Milne-Thomson’s method,

where c is an arbitrary complex constant.

Example 2.52

If v(x,y) = x4 – 6x2y2 + y4 find f (z) = u(x, y) + iv(x, y such that f (z) is analytic.

Solution    Differentiating v partially w.r.t x and y twice, we get

 vx = 4x3 − 12xy2 = −uy (1) vy = −12x2y + 4y3 = ux (2) vxx = 12x2 − 12y2) (3) vyy = −12x2 + 12y2 (4)

⇒ ∇2v = 0 ⇒ v is harmonic

Integrating Eq. (2) w.r.t x, we get

u = −4x3y + 4xy3 + A(y)            (5)

where A(y) is an arbitrary function of y Differentiating w.r.t to y we get

 uy = −4x3 + 12xy2 + A′(y) = −vx = −4x3 + 12xy2 ⇒ A′(y) = 0 A(y) = c =   constant ∴ u(x,y) = −4x3y + 4xy3 + c Hence f(z) = (−4x3y + 4xy3 + c) + i (x4 − 6x2y2 + y4) = i[(x4 − 6x2y2 + y4) + i(4x3y − 4xy3)] + c = i(x + iy)4 + c = iz4 + c

Milne-Thomson’s method:

 Let   Ψ1 (x,y) = vy = −12x2y + 4y3 and   Ψ2x,y = vx = 4x3 − 12xy2 Ψ1 (z, 0) = 0 and Ψ2 (z,0) = 4z3

where c is an arbitrary complex constant.

Example 2.53

Show that u = sin x cosh y + 2 cos x sinh y + x2y2 + 4xy satisfies the Laplace equation. Find the corresponding analytic function.

[JNTU 2008]

Solution    It is given that

u(x,y) = sin xcosh y + 2 cos x sinh y + x2y2 + 4xy       (1)

Differentiating Eq. (1) twice partially w.r.t x and y, respectively, we get

 ux = cos x cosh y − 2 sin x sinh y + 2x + 4y (2) uxx = − sin x cosh y − 2 cos x sinh y + 2 (3) uy = sin x sinh y + 2 cos x cosh y − 2y + 4x (4) uyy = sin x cosh y + 2 cos x sinh y − 2 (5)

Adding Eqs. (3) and (5), we obtain

2u = uxx + uyy = 0                 (6)

u satisfies Laplace’s equation and hence u is a harmonic function.

Also,

Replace x by z and y by 0 (Milne-Thomson’s method)

⇒ cos x cosh y − 2 sin x sinh y + 2x + 4yi(sin x sinh y + 2 cos x cosh y − 2y + 4x)

= Φ1(z, 0) − iΦ2(z, 0) = Φ(z)    (say)

= (cos z · 1 − 0 + 2z + 0) − i(sin z · 0 + 2 cos z · 1 − 2 × 0 + 4z)

Or    Φ(z) = (1 − 2i) cos z + 2(1 − 2i)z

Integrating w.r.t z,

Example 2.54

Find the analytic function whose imaginary part is e–x (x cos y + y sin y).

[JNTU 1998S]

Solution    Let f(z) = u + iv be the analytic function. Then v = e–x (x cos y + y sin y)

 We know that f′(z) = ux + ivx = vy + ivx = Φ1 + iΦ2 = Φ(z)    say Since f (z) = u + iv ∈ H(D), ux = uy, uy, = −vx (CREs) Now vx = e−x (−x cos y − y sin y + cos, y = Φ2(z) = e−z (−z + 1) vy = e−x(−x sin y + sin y + y cos y) = Φ1(z) = e−z (0) Φ(z) = Φ1(z,0) + iΦ2(z,0) = i(i −z)e−z

Integrating w.r.t z, we get

Example 2.55

Find the analytic function whose real part is u = ex [(x2y2) cos y – 2xy sin y].

[JNTU 2008]

Solution

 u = ex[(x2 − y2) cos y − 2xy sin y] ⇒ ux = ex[(x2 − y2) cos y − 2xy sin y + 2x cos y − 2y sin y] = Φ1 (x, y) uy = ex[−2y cos y − (x2 − y2) sin y − 2x sin y −2xy cos y] = Φ2(x, y) ∴ f′ (z) = ux + ivx = ux − iuy = Φ1(z, 0) − iΦ2(z, 0) = ez (z2 + 2z) − ez (0) = ez (z2 + 2z)

Integrating, f (z) = z2ez + c

Example 2.56

If f (z) = u + ivH (D) then prove that

1. 2|f (z)|2 = 4|f′(z)|2.

JNTU 1993]

2. 2up = p(p − 1)up−2|f′(z)|2.

[JNTU 1997S]

Solution    Since f (z) = v + iv is analytic in a domain D, the functions u = u(x, y) and v = v(x,y) satisfy Cauchy-Riemann equations in D

ux = vy     uy = −vx                 (1)

Further, u and v are harmonic functions in D which implies that u, v satisfy Laplace’s equation ∇2 Φ = 0, i.e., ∇2u = 0, ∇2u = 0 in D

1. f (z) = u + iv ⇒ |f (z)|2 = u2 + v2

We calculate partial derivatives w.r.t x and y of |f (z)|2 = u2 + v2

Similarly,

Adding the two results (3) and (4)

 = 4|ux − iuy|2 = 4|f′(z)|2 (5)

2. Differentiating up partially w.r.t x twice, we get

Similarly, we obtain

Adding Eqs. (6) and (7), we obtain

##### EXERCISE 2.3
1. Prove that the following functions are harmonic. Find their harmonic conjugates.
1. u = x2y2y,

Ans: v = 2xy + x + c.

2. Ans:

3. u = 3xy2x3,

Ans: v = y3 – 3x2y + c

4. v = y2x2,

Ans: u = 2xy + c

2. Find the analytic function f (z) =u + iv in question 1.

Ans:

1. z2 + iz + c
2. z3 + ic
3. iz2 + c
3. Find the analytic function f (z) = v + iv such that and f (π / 2) = (3 − i) / 2.

Ans: f (z) = cot(z/2) + (1 − i)/2

4. Find the analytic function f (z) = u + iv such that and f (π / 2) = 0.

Ans:

5. Find the analytic function f (z) = u + iv such that

Ans:

6. If f (z) = u + iv is an analytic function of z = xiy and uv = cx (cos y − sin y) find f (z) in terms of z.

[JNTU 1995S]

Ans: f (z) = ez + c(1 + i)

7. Show that u(x, y)= sin x cosh y + 2 cos x sinh y + x2y2 + 4xy is harmonic. Find an analytic function f (z) with u as its real point.

Ans: f (z) = sin z + z2 − 2i sin z − 2iz2 + c

8. Find the real part of the analytic function whose imaginary part is e-x[2xy cos y + (y2x2) sin y]. Find the analytic function.

Ans: u = e−x[(x2y2) cos y + 2xy sin y]

f (z) = e−z · z2

9. Prove that u + iv is analytic if u = Φy − ψx and v = Φx − ψy where Φ and ψ are harmonic function.
10. If f (z) is analytic, prove that
11. If f(z) = u + iv is analytic prove that and ∇2(amp f (z)) = 0.
12. From Laplace’s equation for u(x, y) prove that .
13. Show that f (x, y) = x3yxy3 + xy + x + y can be the imaginary part of an analytic function of z = x + iy.
14. Find the analytic function f (z) = u(r, θ) + iv(r, θ) such that
1. u(r, θ) = −r3 sin 3θ
2. .

Ans:

1. iz3 + ic
15. If show that the curves u(x, y) = c1 and v(x, y) = c2 intersect orthogonally.